文章目錄
- 題目:[P3372 【模板】線段樹 1](https://www.luogu.org/problemnew/show/P3372)
- 題目:[P3373 【模板】線段樹 2](https://www.luogu.org/problemnew/show/P3373)
- 題目:[I Hate It](https://vjudge.net/problem/HDU-1754)
- 題目:[Mayor's posters](https://vjudge.net/problem/POJ-2528)
- 題目:[Balanced Lineup](https://vjudge.net/problem/POJ-3264)
- 題目:[Can you answer these queries?](https://vjudge.net/problem/HDU-4027)
- 題目:[Tunnel Warfare](https://vjudge.net/problem/HDU-1540)
- 題目:[Assign the task](https://vjudge.net/problem/HDU-3974)
參考博客:
基本線段樹模板(建樹、點/區間修改、查詢)
題目:P3372 【模板】線段樹 1
代碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
#define LL long long
LL a[maxn<<2], st[maxn<<2], add[maxn<<2];
void build(LL o, LL l, LL r){
if(l == r)st[o] = a[l];
else {
LL m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = st[o<<1] + st[o<<1|1];
}
}
void pushup(LL o){
st[o] = st[o<<1] + st[o<<1|1];
}
void pushdown(LL o, LL l, LL r){
if(add[o]){
add[o<<1] += add[o];
add[o<<1|1] += add[o];
LL m = l + ((r-l)>>1);
st[o<<1] += add[o]*(m-l+1);
st[o<<1|1] += add[o]*(r-m);
add[o] = 0;
}
}
void update(LL o, LL l, LL r, LL ql, LL qr, LL addv){
if(ql <= l && qr >= r){
add[o] += addv;
st[o] += addv*(r-l+1);
return;
}
pushdown(o, l, r);
LL m = l + ((r-l)>>1);
if(ql <= m)update(o<<1, l, m, ql, qr, addv);
if(qr >= m+1)update(o<<1|1, m+1, r, ql, qr, addv);
pushup(o);
}
LL query(LL o, LL l, LL r, LL ql, LL qr){
if(ql<= l && qr >= r)return st[o];
pushdown(o, l, r);
LL m = l + ((r-l)>>1);
LL ans = 0;
if(ql <= m)ans += query(o<<1, l, m, ql, qr);
if(qr >= m+1)ans += query(o<<1|1, m+1, r, ql, qr);
return ans;
}
int main(){
LL n, m;
scanf("%lld %lld", &n, &m);
for(LL i = 1; i <= n; i++)scanf("%lld", &a[i]);
build(1, 1, n);
while(m--){
LL a, b, c, d;
scanf("%lld", &a);
if(a == 1){
scanf("%lld %lld %lld", &b, &c, &d);
update(1,1, n, b, c, d);
}
else {
scanf("%lld %lld", &b, &c);
printf("%lld\n", query(1,1, n, b, c));
}
}
return 0;
}
題目:P3373 【模板】線段樹 2
代碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
#define LL long long
LL a[maxn<<2], st[maxn<<2], add[maxn<<2], mul[maxn<<2];
LL mod;
void build(LL o, LL l, LL r)
{
mul[o] = 1;
add[o] = 0;
if(l == r)st[o] = a[l];
else
{
LL m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = (st[o<<1]+st[o<<1|1]) % mod;
}
}
void pushup(LL o)
{
st[o] = (st[o<<1] + st[o<<1|1]) % mod;
}
void pushdown(LL o, LL l, LL r)
{
LL m = l + ((r-l)>>1);
st[o<<1] = (st[o<<1]*mul[o]+add[o]*(m-l+1))%mod;
st[o<<1|1] = (st[o<<1|1]*mul[o]+add[o]*(r-m))%mod;
mul[o<<1] = (mul[o]*mul[o<<1])%mod;
mul[o<<1|1] = (mul[o<<1|1]*mul[o])%mod;
add[o<<1] = (add[o<<1]*mul[o]+add[o])%mod;
add[o<<1|1] = (add[o<<1|1]*mul[o] + add[o])%mod;
add[o] = 0;
mul[o] = 1;
return;
//}
}
void update(LL o, LL l, LL r, LL ql, LL qr, LL addv)
{
if(ql <= l && qr >= r)
{
add[o] =(add[o] + addv) % mod;
st[o] =(st[o] + addv*(r-l+1))%mod;
return;
}
pushdown(o, l, r);
LL m = l + ((r-l)>>1);
if(ql <= m)update(o<<1, l, m, ql, qr, addv);
if(qr >= m+1)update(o<<1|1, m+1, r, ql, qr, addv);
pushup(o);
return;
}
void update1(LL o, LL l, LL r, LL ql, LL qr, LL addv)
{
if(ql <= l && qr >= r)
{
add[o] = (add[o]*addv)%mod;
mul[o] = (mul[o]*addv)%mod;
st[o] = (st[o]*addv)%mod;
return;
}
pushdown(o, l, r);
LL m = l + ((r-l)>>1);
if(ql <= m)update1(o<<1, l, m, ql, qr, addv);
if(qr >= m+1)update1(o<<1|1, m+1, r, ql, qr, addv);
pushup(o);
return;
}
LL query(LL o, LL l, LL r, LL ql, LL qr)
{
LL ans = 0;
if(ql<= l && qr >= r)return st[o];
pushdown(o, l, r);
LL m = (l + r)>>1 ;
if(ql <= m)ans = (ans + query(o<<1, l, m, ql, qr))%mod;
if(qr >= m+1)ans = (ans + query(o<<1|1, m+1, r, ql, qr)) % mod;
return ans;
}
int main()
{
LL n, m, x, y, z, op;
scanf("%lld %lld %lld", &n, &m, &mod);
for(LL i = 1; i <= n; i++)scanf("%lld", &a[i]);
build(1, 1, n);
while(m--)
{
scanf("%lld", &op);
if(op == 1)
{
scanf("%lld %lld %lld", &x, &y, &z);
update1(1, 1, n, x, y, z);
}
else if(op == 2)
{
scanf("%lld %lld %lld", &x, &y, &z);
update(1, 1, n, x, y, z);
}
else if(op == 3)
{
scanf("%lld %lld", &x, &y);
printf("%lld\n", query(1, 1, n, x, y));
}
else
{
for(int i = 1; i <= n; i++)
{
printf("%lld ", query(1, 1, n, i, i));
}
printf("\n");
}
}
}
題目:I Hate It
代碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int st[maxn<<2];
void build(int o, int l, int r)
{
if(l == r)scanf("%d", &st[o]);
else
{
int m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = max(st[o<<1], st[o<<1|1]);
}
}
void update(int o, int l, int r, int ind, int ans)
{
if(l == r)
{
st[o] = ans;
return;
}
int m = l + ((r-l)>>1);
if(ind <= m)update(o<<1, l, m, ind, ans);
else update(o<<1|1, m+1, r, ind, ans);
st[o] = max(st[o<<1], st[o<<1|1]);
}
int query(int o, int l, int r, int ql, int qr)
{
if(ql > r || qr < l)return -1;
if(ql <= l && qr >= r)return st[o];
int m = l + ((r-l)>>1);
int p1 = query(o<<1, l, m, ql, qr), p2 = query(o<<1|1, m+1, r, ql, qr);
return max(p1, p2);
}
int main()
{
int n, m, x, y, z;
char op;
while( ~scanf("%d %d", &n, &m))
{
build(1, 1, n);
while(m--)
{
getchar();
scanf("%c", &op);
if(op == 'Q')
{
scanf("%d %d", &x, &y);
printf("%d\n", query(1, 1, n, x, y));
}
else
{
scanf("%d %d", &x, &y);
update(1, 1, n, x, y);
}
}
}
}
題目:Mayor’s posters
題意:就是告訴你,有一面牆,牆的寬度是1e7,然後讓你貼海報,海報的高度是跟牆的高度是一致的,每次粘貼海報會把之前的海報覆蓋掉,問你最後能看見的海報有幾張,只要這個海報沒有完全被覆蓋掉,那麼這個海報就能露出來。
思路:本體的思路,因爲涉及到區間修改,所以我就想到了線段樹,線段樹的區間修改。但是如何去查詢海報剩下幾張呢,我們把海報做成一個編號,每次的區間覆蓋,我們就把這個區間的值賦值爲這個海報的編號,然後通過一個數組去標記,然後統計有多少個數組已經被標記了。
但是存在一個問題,數據的大小太大了,但是數據的個數不大,所以很容易想到離散化,通過離散化把數據變得小一點。
代碼:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 1e5+10;
int st[maxn<<2], add[maxn<<2];
struct Node
{
int x, y;
} a[maxn];
int vis[maxn], b[maxn];
void build(int o, int l, int r)
{
if(l == r)st[o] = 0;
else
{
int m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = st[o<<1] + st[o<<1|1];
}
}
void pushup(int o)
{
st[o]=st[o<<1]+st[o<<1|1];
}
void pushdown(int o,int l,int r) //pushdown和區間加值不同,改值時修改結點信息只需要對修改後的信息求和即可,不用加上原信息
{
if(add[o])
{
int c=add[o];
add[o<<1]=c;
add[o<<1|1]=c;
int m=l+((r-l)>>1);
st[o<<1]=(m-l+1)*c;
st[o<<1|1]=(r-m)*c;
add[o]=0;
}
}
void update(int o,int l,int r,int ql,int qr,int c)
{
if(ql<=l&&qr>=r) //同樣更新結點信息和區間加值不同
{
add[o]=c;
st[o]=(r-l+1)*c;
return;
}
pushdown(o,l,r);
int m=l+((r-l)>>1);
if(ql<=m)update(o<<1,l,m,ql,qr,c);
if(qr>=m+1)update(o<<1|1,m+1,r,ql,qr,c);
pushup(o);
}
int query(int o,int l,int r,int ql,int qr)
{
if(ql<=l&&qr>=r) return st[o];
pushdown(o,l,r);
int m=l+((r-l)>>1);
int ans=0;
if(ql<=m)ans+=query(o<<1,l,m,ql,qr);
if(qr>=m+1)ans+=query(o<<1|1,m+1,r,ql,qr);
return ans;
}
int main()
{
int T, n;
scanf("%d", &T);
while(T--)
{
memset(vis, 0, sizeof(vis));
scanf("%d", &n);
int tot = 0;
for(int i = 1; i <= n; i++)
{
scanf("%d %d", &a[i].x, &a[i].y);
b[tot++] = a[i].x;
b[tot++] = a[i].y;
}
sort(b, b+tot);
for(int i = 1; i <= n; i++)
{
a[i].x = lower_bound(b,b+tot, a[i].x)-b+1;//離散化
a[i].y = lower_bound(b,b+tot, a[i].y)-b+1;
}
build(1, 1, tot);
for(int i = 1; i <= n; i++)update(1, 1, tot, a[i].x, a[i].y, i);
for(int i = 1; i <= tot; i++)
{
vis[query(1, 1, tot, i, i)] = 1;//標記
}
int cn = 0;
for(int i = 1; i <= tot; i++)
{
if(vis[i])cn++;
}
printf("%d\n", cn);
}
return 0;
}
題目:Balanced Lineup
題意:本題的題意就是給你一個區間求區間的最大值和最小值的差值。
思路:建兩個線段樹,一個用來維護最大值,另外一個用來維護最小值。
代碼:
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int INF = 2e5+10;
int st[INF<<2], b[INF<<2];
int tt[INF<<2], a[INF<<2];
const int maxn = 1e7;
void build(int o, int l, int r)
{
if(l == r)st[o] = a[l];
else
{
int m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = max(st[o<<1],st[o<<1|1]);
}
}
void build1(int o, int l, int r)
{
if(l == r)tt[o] = a[l];
else
{
int m = l + ((r-l)>>1);
build1(o<<1, l, m);
build1(o<<1|1, m+1, r);
tt[o] = min(tt[o<<1],tt[o<<1|1]);
}
}
int query(int o,int l,int r,int ql,int qr)
{
if(ql<=l&&qr>=r) return st[o];
int m=l+((r-l)>>1);
int ans1 = 0, ans2 = 0;
if(ql<=m)ans1 = query(o<<1,l,m,ql,qr);
if(qr>=m+1)ans2 = query(o<<1|1,m+1,r,ql,qr);
return max(ans1, ans2);
}
int query1(int o,int l,int r,int ql,int qr)
{
if(ql<=l&&qr>=r) return tt[o];
int m=l+((r-l)>>1);
int ans1=maxn, ans2 = maxn;
if(ql<=m)ans1 = query1(o<<1,l,m,ql,qr);
if(qr>=m+1)ans2 = query1(o<<1|1,m+1,r,ql,qr);
return min(ans1, ans2);
}
int main()
{
int n, m;
scanf("%d %d", &n, &m);
for(int i = 1; i <= n; i++)scanf("%d", &a[i]);
for(int i = 1; i <= n; i++)
{
tt[i] = maxn;
}
build(1, 1, n);
build1(1, 1, n);
for(int i = 1; i <= m; i++)
{
int l, r;
scanf("%d %d", &l, &r);
int res = query(1, 1, n, l, r)- query1(1, 1, n, l, r);
printf("%d\n", res);
}
}
題目:Can you answer these queries?
題意:本題的題意是把區間的每一個數開根號,然後進行區間查詢。
思路:本題的思路就是線段樹的區間修改區間查詢,但是這個修改,需要一些技巧,就是當這個數的區間的所有數都是1的情況下就不需要更新了,這麼就不會超時了。
代碼:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+10;
#define LL long long
LL st[maxn<<2], a[maxn<<2];
void build(int o, int l, int r)
{
if(l == r)st[o] = a[l];
else
{
int m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = st[o<<1] + st[o<<1|1];
}
}
void update(int o,int l,int r,int ql, int qr)
{
if(l==r) //若當前更新點的左右端點相等即到葉子結點時,直接更新信息並返回
{
st[o] = sqrt(st[o]);
return;
}
if(ql<=l&&qr>=r&&st[o]==r-l+1)//如果區間內的所有數都是1則不必更新
return;
int m = l+((r-l)>>1);
if(ql<=m)
update(o<<1,l,m, ql, qr);
if(m+1<=qr)
update((o<<1)|1,m+1,r, ql, qr);
st[o] = st[o<<1] + st[(o<<1)|1];//遞歸回之後用兒子結點更新父節點(此處是區間最大值)
}
LL query(int o,int l,int r,int ql,int qr){
if(ql<=l&&qr>=r) return st[o];
int m=l+((r-l)>>1);
LL ans=0;//所需查詢的結果
if(ql<=m)ans+=query(o<<1,l,m,ql,qr);
if(qr>=m+1)ans+=query(o<<1|1,m+1,r,ql,qr);
return ans;
}
int main()
{
int n, m;
int ca = 0;
while(~scanf("%d", &n)){
for(int i = 1; i <= n; i++)scanf("%lld", &a[i]);
build(1, 1, n);
scanf("%d", &m);
printf("Case #%d:\n",++ca);
while(m--)
{
int op, l, r;
scanf("%d %d %d", &op, &l, &r);
if(l > r)swap(l, r);//z注意題目中沒有限制l, r的大小
if(op == 0)
{
update(1, 1, n, l, r);
}
else printf("%lld\n", query(1, 1, n, l, r));
}
printf("\n");
}
return 0;
}
題目:Tunnel Warfare
題意:本題的題意很清楚,就是給一一條線段,這個線段中連接了n個村莊,然後有m個操作,D是把某個村轉摧毀,R是把最近摧毀的村莊修復,注意如果最新的村莊已經修復完畢,那麼下一次修復就是上一個被摧毀的,Q是查詢有多少個村莊與該村莊相連接。
思路:本題的思路就是用線段樹用來維護區間,如果這個村莊被摧毀那麼這個位置就是0,否則就是1,然後在通過過二分查找來確定範圍,注意多組輸入,和不能用cin和cout.
代碼:
#include<iostream>
#include<cmath>
#include<stack>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 5e4+10;
stack<int>s;
int st[maxn<<2];
void build(int o, int l, int r)
{
if(l == r)st[o] = 1;
else
{
int m = l + ((r-l)>>1);
build(o<<1, l, m);
build(o<<1|1, m+1, r);
st[o] = st[o<<1] + st[o<<1|1];
}
}
void update(int o, int l, int r, int ind, int ans)
{
if(l == r)
{
st[o] = ans;
return;
}
int m = l+((r-l)>>1);
if(ind <= m)update(o<<1, l, m, ind, ans);
else update(o<<1|1, m+1, r, ind, ans);
st[o] = st[o<<1]+st[o<<1|1];
}
int query(int o,int l,int r,int ql,int qr)
{
if(ql<=l&&qr>=r) return st[o];
int m=l+((r-l)>>1);
int ans=0;
if(ql<=m)ans+=query(o<<1,l,m,ql,qr);
if(qr>=m+1)ans+=query(o<<1|1,m+1,r,ql,qr);
return ans;
}
bool check(int mid, int pos, int n)
{
if(query(1, 1, n, pos, mid) == mid-pos+1)return true;
else return false;
}
int bsearch_2(int l, int r, int pos, int n)
{
while (l < r)
{
int mid = l + r + 1 >> 1;
if (check(mid, pos, n)) l = mid;
else r = mid - 1;
}
return l;
}
bool check1(int mid, int pos, int n)
{
if(query(1, 1, n, mid, pos) == pos-mid+1)return true;
else return false;
}
int bsearch_1(int l, int r, int pos, int n)
{
while (l < r)
{
int mid = l + r >> 1;
if (check1(mid, pos, n)) r = mid;
else l = mid + 1;
}
return l;
}
int main()
{
int n, m;
while(scanf("%d %d", &n, &m)!=EOF)
{
build(1, 1, n);
char op;
while(m--)
{
int pos, pre;
getchar();
scanf("%c", &op);
if(op == 'D')
{
cin>>pos;
update(1, 1, n, pos, 0);
s.push(pos);
}
else if(op == 'R')
{
if(s.empty())continue;
update(1, 1, n, s.top(), 1);
s.pop();
}
else
{
scanf("%d", &pos);
int t = query(1, 1, n, pos, pos);
if(t == 0)printf("0\n");
else
{
int v_r = bsearch_2(pos, n, pos, n);
int v_l = bsearch_1(1, pos, pos,n);
printf("%d\n", v_r-v_l+1);
}
}
}
}
}
題目:Assign the task
題意:題意就是給你一棵樹,每次修改一個子書的所有結點,然後查詢某個結點的值。
思路:本題看見修改和查詢很容易想到線段樹,但是這個樹的子樹數如果每次都搜索一遍,那麼一定是非常耗時的,所以我當時就想到樹鏈剖分,不知道是爆棧了還是我沒有用明白,結果一直RE,所以我就換了一個思路採用dfs序的方法,把這棵樹的所有結點重新設置一個序列,然後利用這個序列進行區間修改單點查詢。
代碼:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mid l+((r-l)>>1)
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
#define len (r-l+1)
#define Temp template<typename T>
#define len (r-l+1)
#define LL long long
Temp inline void read(T &x)
{
x = 0;
T w = 1, ch = getchar();
while(!isdigit(ch) && ch!= '-')ch = getchar();
if(ch == '-')w = -1, ch = getchar();
while(isdigit(ch))x = (x<<3)+(x<<1)+(ch^'0'),ch = getchar();
x = x * w;
}
const int maxn = 300000+10;
int n, q, r;
int e, beg[maxn<<1], nex[maxn<<1], to[maxn<<1], wt[maxn<<1], degree[maxn<<1];
LL a[maxn<<2];
int laz[maxn<<2];
int s[maxn], ex[maxn], cnt = 0 ;
LL res = 0;
inline void add(int x, int y)
{
to[++e] = y;
nex[e] = beg[x];
beg[x] = e;
}
inline void pushdown(int rt, int l, int r)
{
laz[rt<<1] = laz[rt];
laz[rt<<1|1] = laz[rt];
int m = l + ((r-l)>>1);
a[rt<<1] = laz[rt]*(m-l+1);
a[rt<<1|1] = laz[rt]*(r-m);
laz[rt] = -1;
}
inline void build(int rt, int l, int r)
{
if(l == r)
{
a[rt] = -1;
return;
}
build(lson);
build(rson);
//a[rt] = (a[rt<<1]+a[rt<<1|1]);
}
inline void query(int rt, int l, int r, int L, int R)
{
if(L <= l && r <= R)
{
res = a[rt];
return;
}
else
{
if(laz[rt] != -1)pushdown(rt, l, r);
if(L <= mid)query(lson, L, R);
if(R > mid)query(rson, L, R);
}
}
inline void update(int rt, int l, int r, int L, int R, int k)
{
if(L <= l && r<= R)
{
laz[rt] = k;
a[rt] = k * len;
return;
}
else
{
if(laz[rt] != -1)pushdown(rt, l, r);
if(L <= mid)update(lson, L, R, k);
if(R > mid)update(rson, L, R, k);
//a[rt] = (a[rt<<1]+a[rt<<1|1]);
}
}
inline LL qson(int x)
{
res = 0;
query(1, 1, n, s[x], s[x]);
return res;
}
inline void updson(int x, int k)
{
update(1, 1, n,s[x], ex[x], k);
}
void dfs(int u){
cnt++;
s[u] = cnt;
for(int i = beg[u]; i; i = nex[i]){
dfs(to[i]);
}
ex[u] = cnt;
}
int main()
{
int T;
int ca = 0;
read(T);
while(T--)
{
memset(degree, 0, sizeof degree);
memset(laz, -1, sizeof(laz));
memset(beg, 0, sizeof beg);
e = 0;
cnt = 0;
read(n);
for(int i = 1; i < n; i++)
{
int a, b;
read(a);
read(b);
add(b, a);
degree[a]++;
}
for(int i = 1; i <= n; i++){
if(degree[i] == 0){
r = i;
break;
}
}
dfs(r);
build(1, 1, n);
read(q);
printf("Case #%d:\n", ++ca);
while(q--){
char op;
int root, val;
cin>>op;
if(op == 'C'){
read(root);
res = 0;
printf("%lld\n",qson(root));
}
else {
read(root);
read(val);
updson(root, val);
}
}
}
return 0;
}