澱粉質(點分治)簡單食用指南

第一次學點分治是去年的九月份。
到現在已經一年了。
今天希望通過這篇博客讓大家都能好好吸收澱粉質。
首先我們要知道分治是什麼
天下大勢 分久必合 合久必分
什麼是合？什麼是分？
給你一棵樹長這樣
我們發現上面有重心兩個字 重心真是一個好的性質 關於爲什麼選重心可以觀看 漆子超的論文
這樣我們對第一個重心處理出所有路徑 然後發現 路徑有圖上六條
我們需要兩兩合併 就像A - D 和 A - B合併 就變成B -D
但是大家返現A - B -C 和 A - B -E合併就會有重邊
在我們統計一個樹上路徑爲k的時候這樣顯然會出問題
我們想一下是哪裏出問題了？
就是子樹裏的兩兩匹配的時候顯然會出錯
所以我們對子樹也要進行一次操作 只不過這次操作是減去他的影響
比如 你看你到B的時候 變成B , B - C ,B - E
但是他們合併好像是正常的？注意我們solve函數 我們會把上條邊傳進去在這個例子也就是A 到 B
的邊
下面是幾道例題
POJ1741
小於等於k的路徑數 排序後雙指針即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 20025;
int eid,p[MAX_N],k,Size[MAX_N],maxson[MAX_N],SIZE,ms,root,now[MAX_N],dep[MAX_N],ans;
bool vis[MAX_N];
struct node
{
    int next,v,dis;
}e[MAX_N<<1];
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
void add(int u,int v,int dis)
{
    e[eid].v = v;
    e[eid].dis = dis;
    e[eid].next = p[u];
    p[u] = eid++;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]||v==fa) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[v]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE-Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getdis(int x,int fa)
{
    now[++now[0]] = dep[x];
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        dep[v]=dep[x]+e[i].dis;
        getdis(v,x);
    }
}
int solve(int x,int dis,int flag)
{
    dep[x] = dis;now[0] = 0;getdis(x,-1);
    sort(now+1,now+1+now[0]);
    int L = 1,R = now[0],sum = 0;
    while(L<R) if(now[L]+now[R]<=k) sum+=R-L,L++;else R--;
    return sum;
}
void fenzhi(int x,int ssize)
{
    vis[x] = true;
    ans+=solve(x,0,0);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        ans-=solve(v,e[i].dis,1);
        ms = inf,root = 0;
        SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);
        getroot(v,0);
        fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,a,b,c;
    while(scanf("%d%d",&n,&k)==2&&(n|k))
    {
        memset(vis,false,sizeof(bool)*(n+1));
        ans = 0;
        init();
        for(int i = 1;i<n;++i)
        {
            scanf("%d%d%d",&a,&b,&c);
            add(a,b,c);add(b,a,c);
        }
        root = 0;ms = inf;SIZE = n;
        getroot(1,-1);
        fenzhi(root,n);
        printf("%d\n",ans);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

P3806
問路徑和爲k的路徑是否存在
發現詢問k不大 用桶標記 O1輸出

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 10025;
int p[MAX_N],eid,SIZE,root,ms,dep[MAX_N],Size[MAX_N],maxson[MAX_N],now[MAX_N];
int cnt[10000025];
bool vis[MAX_N];
struct edge
{
    int v,next,dis;
}e[MAX_N<<1];
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
void add(int u,int v ,int dis)
{
    e[eid].v = v;
    e[eid].next = p[u];
    e[eid].dis = dis;
    p[u] = eid++;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]||v==fa) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[v]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE - Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getdis(int x,int fa)
{
    now[++now[0]] = dep[x];
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        dep[v] = dep[x]+e[i].dis;
        getdis(v,x);
    }
}
void solve(int x,int dis,int flag)
{
    dep[x] = dis;now[0]=0;getdis(x,-1);
    sort(now+1,now+1+now[0]);
    int L = 1,R = now[0];
    if(!flag) while(L<R) if(now[L]+now[R]<=10000000) cnt[now[L]+now[R]]++,L++;else R--;
    if(flag) while(L<R) if(now[L]+now[R]<=10000000) cnt[now[L]+now[R]]--,L++;else R--;
}
void fenzhi(int x,int ssize)
{
    vis[x] = true;
    solve(x,0,0);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        solve(v,e[i].dis,1);
        ms = inf;root = 0;
        SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);
        getroot(v,0);
        fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,m,a,b,c,k;
    scanf("%d%d",&n,&m);
    init();
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);add(b,a,c);
    }
    root = 0;ms = inf;SIZE = n;
    getroot(1,-1);
    fenzhi(root,n);
    while(m--)
    {
        scanf("%d",&k);
        if(cnt[k]) printf("AYE\n");
        else printf("NAY\n");
    }

    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

P2634
路徑和爲3
用路徑取模後爲 0 爲 1 爲 2的桶統計答案即可

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 20025;
int p[MAX_N],eid,T[5],Size[MAX_N],maxson[MAX_N],SIZE,root,ms,dep[MAX_N],ans;
bool vis[MAX_N];
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
struct edge
{
    int v,next,dis;
}e[MAX_N<<1];
void add(int u,int v,int dis)
{
    e[eid].v = v;
    e[eid].dis = dis;
    e[eid].next = p[u];
    p[u] = eid++;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[x]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE - Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getdis(int x,int fa)
{
    dep[x]%=3;
    T[dep[x]]++;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        dep[v] = (dep[x]+e[i].dis)%3;
        getdis(v,x);
    }
}
int solve(int x,int dis)
{
    dep[x] = dis;T[0] = T[1] = T[2] = 0;getdis(x,-1);
    return T[0]*T[0]+T[1]*T[2]*2;
}
void fenzhi(int x,int ssize)
{
    vis[x] = true;
    ans+=solve(x,0);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
         ans-=solve(v,e[i].dis);
         ms = inf;root = 0;
         SIZE = Size[v]<Size[x]?Size[v]:(ssize - Size[x]);
         getroot(v,0);
         fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,a,b,c;
    scanf("%d",&n);
    init();
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d%d",&a,&b,&c);
        add(a,b,c);add(b,a,c);
    }
    ms = inf;root = 0;SIZE = n;
    getroot(1,-1);
    fenzhi(root,n);
    int down = n*n;
    int GCD = gcd(ans,down);
    printf("%d/%d",ans/GCD,down/GCD);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

P4149
只要加邊的時候賦值
刪邊的時候賦無窮大(1條邊肯定比組合起來更優
所以可以直接刪

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 200025;
int p[MAX_N],eid,k,dep[MAX_N],Size[MAX_N],SIZE,ms,root,maxson[MAX_N],ans;
bool vis[MAX_N];
int cnt[1000025];
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
struct edge
{
    int v,next,dis;
}e[MAX_N<<1];
void add(int u,int v,int dis)
{
    e[eid].v = v;
    e[eid].next = p[u];
    e[eid].dis = dis;
    p[u] = eid++;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[v]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE-Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getans(int x,int fa,int dis,int now)
{
    if(dis>k) return;
    ans = min(ans,cnt[k-dis]+now);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        getans(v,x,dis+e[i].dis,now+1);
    }

}
void update(int x,int fa,int dis,int now)
{
    if(dis>k) return;
    cnt[dis] = min(cnt[dis],now);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        update(v,x,dis+e[i].dis,now+1);
    }
}
void Clear(int x,int fa,int dis)
{
    if(dis>k) return;
    cnt[dis] = 1e9;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        Clear(v,x,dis+e[i].dis);
    }
}
void fenzhi(int x,int ssize)
{
    cnt[0] = 0;
    vis[x] = true;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        getans(v,x,e[i].dis,1);
        update(v,x,e[i].dis,1);
    }
    Clear(x,0,0);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;if(vis[v]) continue;
        ms = inf;root = 0;
        SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);
        getroot(v,0);
        fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,a,b,c;
    scanf("%d%d",&n,&k);
    init();
    ans = 1e9;
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d%d",&a,&b,&c);a++,b++;
        add(a,b,c);add(b,a,c);
    }
    for(int i = 0;i<=k;++i) cnt[i] = 1e9;
    ms = inf;root = 0;
    SIZE = n;
    getroot(1,-1);
    fenzhi(root,n);
    if(ans>=1e9) printf("-1\n");
    else printf("%d\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

cf293E
發現長度不大 可以對長度建樹狀數組 每次在區間內查詢滿足的個數即可

//現在有一棵樹，每條邊的長度都爲1，然後有一個權值，求存在多少個(u,v)點對，他們的路勁長度 <= l, 總權重 <= w.
/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 100025;
int C[MAX_N<<1];
void ADD(int x,int v)
{
    for(;x<MAX_N;x+=x&(-x))
        C[x]+=v;
}
int getsum(int x)
{
    int res = 0;
    if(x<=0) return 0;
    for(;x;x-=x&(-x))
        res+=C[x];
    return res;
}
int n,l,W,p[MAX_N],eid,Size[MAX_N],SIZE,ms,root,maxson[MAX_N],dep[MAX_N],weight[MAX_N];
ll ans;
bool vis[MAX_N];
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
struct edge
{
    int v,next,dis,w;
}e[MAX_N<<1];
struct node
{
    int a,b;
    bool operator < (const node other) const
    {
        return b < other.b;
    }
}now[MAX_N];
void add(int u,int v,int dis,int w)
{
    e[eid].v = v;
    e[eid].next = p[u];
    e[eid].dis = dis;
    e[eid].w = w;
    p[u] = eid++;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]||v==fa) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[v]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE - Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getdis(int x,int fa)
{
    now[++now[0].a].a = dep[x];now[now[0].a].b = weight[x];
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        dep[v] = dep[x] + 1;
        weight[v] = weight[x] + e[i].w;
        getdis(v,x);
    }
}
ll solve(int x,int dis,int w)
{
    dep[x] = dis;weight[x] = w;now[0].a = 0;getdis(x,-1);
    sort(now+1,now+1+now[0].a);
    int L = 1,R = now[0].a;
    ll sum = 0;
    for(int i = 1;i<=now[0].a;++i) ADD(now[i].a+1,1);
    while(L<R)
    {
        if(now[L].b+now[R].b<=W) ADD(now[L].a+1,-1),sum+=1ll*getsum(l-now[L].a+1),L++;
        else ADD(now[R].a+1,-1),R--;
    }
    ADD(now[L].a+1,-1);
    return sum;
}
void fenzhi(int x,int ssize)
{
    vis[x] = true;
    ans+=solve(x,0,0);
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        ans-=solve(v,e[i].dis,e[i].w);
        ms = inf;root = 0;
        SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);
        getroot(v,0);
        fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int a,b;
    scanf("%d%d%d",&n,&l,&W);
    init();
    for(int i = 2;i<=n;++i)
    {
        scanf("%d%d",&a,&b);
        add(a,i,1,b);add(i,a,1,b);
    }
    ms = inf;root = 0;SIZE = n;
    getroot(1,-1);
    fenzhi(root,SIZE);
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

CF161D
統計路徑和爲k的個數
暴力統計即可 注意我們不想要子樹當前內的和 所以用一個tmp和一個num數組

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
//#pragma comment(linker, "/STACK:10240000,10240000")
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
typedef unsigned long long ull;
const ull hash1 = 201326611;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);
//ull ha[MAX_N],pp[MAX_N];

inline int read()
{
    int date = 0,m = 1; char ch = 0;
    while(ch!='-'&&(ch<'0'|ch>'9'))ch = getchar();
    if(ch=='-'){m = -1; ch = getchar();}
    while(ch>='0' && ch<='9')
    {
        date = date*10+ch-'0';
        ch = getchar();
    }return date*m;
}

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/

/*int root[MAX_N],cnt,sz;
namespace hjt
{
    #define mid ((l+r)>>1)
    struct node{int l,r,maxx;}T[MAX_N*40];

    #undef mid
}*/
const int MAX_N = 50025;
int n,k,p[MAX_N],eid,maxson[MAX_N],dep[MAX_N],root,ms,SIZE,Size[MAX_N],ans;
bool vis[MAX_N];
int num[505],tmp[505];
struct edge
{
    int v,next,dis;
}e[MAX_N<<1];
void add(int u,int v,int dis)
{
    e[eid].dis = dis;
    e[eid].next = p[u];
    e[eid].v = v;
    p[u] = eid++;
}
void init()
{
    memset(p,-1,sizeof(p));
    eid = 0;
}
void getroot(int x,int fa)
{
    Size[x] = 1;maxson[x] = 0;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]||v==fa) continue;
        getroot(v,x);
        Size[x]+=Size[v];
        if(Size[v]>maxson[x]) maxson[x] = Size[v];
    }
    if(SIZE-Size[x]>maxson[x]) maxson[x] = SIZE - Size[x];
    if(ms>maxson[x]) ms = maxson[x],root = x;
}
void getdis(int x,int fa)
{
    if(dep[x]>k) return;
    ans+=num[k-dep[x]];++tmp[dep[x]];
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(v==fa||vis[v]) continue;
        dep[v] = dep[x] + e[i].dis;
        getdis(v,x);
    }
}
void fenzhi(int x,int ssize)
{
    vis[x] = true;
    for(int i = 1;i<=k;++i) num[i] = 0;num[0] = 1;
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        for(int j = 0;j<=k;++j) tmp[j] = 0;
        dep[v] = 1;
        getdis(v,x);
        for(int j = 0;j<=k;++j) num[j] += tmp[j];
    }
    for(int i = p[x];i+1;i=e[i].next)
    {
        int v = e[i].v;
        if(vis[v]) continue;
        ms = inf;root = 0;
        SIZE = Size[v]<Size[x]?Size[v]:(ssize-Size[x]);
        getroot(v,0);
        fenzhi(root,(Size[v]<Size[x]?Size[v]:(ssize-Size[x])));
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int a,b;
    scanf("%d%d",&n,&k);
    init();
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d",&a,&b);
        add(a,b,1);add(b,a,1);
    }
    ms = inf;root = 0;SIZE = n;
    getroot(1,-1);
    fenzhi(root,SIZE);
    printf("%d\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}