2016 UESTC Training for Data Structures (1)

用了1mol多時間+借鑑(chao)了1mol多題解後終於差不多a完了(話說直接把題目copy過來讓我們做真的好?)
A 線段樹模板 不用多說了吧

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define sdrwr 1
using namespace std;
long long n,m,t,a,b;
long long sum[400005];
void pushup(long long rt)
{
    sum[rt]=max(sum[rt<<1],sum[rt<<1|1]);
}
void update(long long rt,long long l,long long r,long long q,long long x)
{
    if (l==r)
    {
        sum[rt]=sum[rt]+x;
        return;
    }
    long long mid=(l+r)>>1;
    if (q<=mid)update(lson,q,x);
    else update(rson,q,x);
    pushup(rt);
}
long long query(long long rt,long long l,long long r,long long x,long long y)
{
    if (l>=x&&r<=y)
    {
        return sum[rt];
    }
    long long mid=(l+r)>>1;
    long long ans=0;
    if (x<=mid)ans=max(ans,query(lson,x,y));
    if (y>mid)ans=max(ans,query(rson,x,y));
    return ans;
}
int main()
{
      scanf("%lld%lld",&n,&m);
    for (long long i=1;i<=m;i++)
    {
        scanf("%lld%lld%lld",&t,&a,&b);
        if (t==1) {update(1,1,n,a,b);}
        if (t==2) {printf("%lld\n",query(1,1,n,a,b));}
    }
    return 0;
}

B 卿學姐與基本法
離散化的線段樹,但我在網上看到了一種神奇的算法

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
map <int,int> q;
map <int,int> ::iterator w;
int t,l,r,x,y,haha;
int query(int x,int y)
{
    int l=x,r=y,ans=y-x+1;
    for (w=q.begin();w!=q.end();w++)
    {
        int a=w->first,b=w->second;
        if (a>r) return ans;
        else if (b<l) continue;
        else if (a<l&&b>r) 
        {
            return ans-=r-l+1;
        }   
        else 
        {
         if(a<=l && b<=r)
                ans-=b-l+1,l=b+1;
            else if(a>l && b<=r)
                ans-=b-a+1,l=b+1;
            else
                return ans-=r-a+1;
        }   
    }
    return ans;
}
int main()
{
    int n,t;
    scanf("%d%d",&n,&t);
    for (int i=1;i<=t;i++)
    {
        scanf("%d%d%d",&x,&l,&r);
        if (x==1)
        {
            if(q[l]<r) q[l]=r;
        }
        else printf("%d\n",query(l,r));
    }
}

C 卿學姐與詭異村莊
並查集,開兩倍內存,如果是一個人另外一個人說是好人,就說明是同類,反之不是.所以如果是一就find(x,y);和find(x+n,y+n);否則find(x,y+n)和find(x+n,y);如果有矛盾就懵逼,否則就行.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
int fa[200005],x,t,n,sdf;
int find(int x)
{
    if (fa[x]==x) return fa[x];
    else return fa[x]=find(fa[x]);
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=2*n;i++)
    fa[i]=i;
    for (int i=1;i<=n;i++)
    {
        scanf("%d%d",&x,&t);
        if (t==1) 
        {
            int x2=find(x);
            int y2=find(i);
            if (x2!=y2)
            {
                fa[x2]=y2;
            }
            x2=find(x+n);
            y2=find(i+n);
            if (x2!=y2)
            {
                fa[x2]=y2;
            }
        }
        else
        {
            int x2=find(x+n);
            int y2=find(i);
            if (x2!=y2)
            {
                fa[x2]=y2;
            }
            x2=find(x);
            y2=find(i+n);
            if (x2!=y2)
            {
                fa[x2]=y2;
            }
        }
    }
    for (int i=1;i<=n;i++)
    {
        if (find(i)==find(i+n))
        {
            printf("One face meng bi");
            return 0;
        }

    }
    printf("Time to show my power");
    return 0;
}

D 卿學姐與魔法
優先隊列,排序,再把所有(a[i]+b[1],1)壓進去就Ok,彈出時把b的下一個壓進去即可,水題一枚

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
int n,a[100005],b[100005],d[100005],ans,cnt=1,min1,min2;
typedef pair<int,int >pii;
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a+1,a+1+n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&b[i]);
    }
    sort(b+1,b+1+n);
    priority_queue<pii, vector<pii>, greater<pii> > q;
    for (int i=1;i<=n;i++)
    {
        q.push(make_pair(a[i]+b[1],1));
    }
    int t=1; 
    for (int i=1;i<=n;i++)
    {
        pii u=q.top();
        q.pop();        
        q.push(make_pair(u.first-b[u.second]+b[u.second+1],u.second+1));
        printf("%d\n",u.first);
    }
    return 0;
}

E 卿學姐與城堡的牆
離散化樹狀數組,蛤蛤蛤,

#include <cstdio>  
#include <cstdlib>  
#include <algorithm>  
using namespace std;  
#define ll long long  
#define maxn 200005  
struct Line  
{  
    ll y1, y2;  
    bool operator < (Line x)const  
    {  
        return y1 < x.y1;  
    }  
}line[maxn];  
int N;  
ll u, v, data[maxn];  
ll tree[maxn];  
void update(int i, ll val)  
{  
    while (i <= N)  
    {  
        tree[i] += val;  
        i += i&(-i);  
    }  
}  
ll query(int i)  
{  
    ll sum = 0;  
    while (i)  
    {  
        sum += tree[i];  
        i -= i&(-i);  
    }  
    return sum;  
}  
int main()  
{   
    scanf("%d%lld%lld", &N, &u, &v);  
    ll a, b;  
    for (int i = 0; i < N; ++i)  
    {  
        scanf("%lld%lld", &a, &b);  
        line[i].y1 = a*u + b, line[i].y2 = a*v + b;  
        data[i] = line[i].y2;  
    }  
    sort(line, line + N);  
    sort(data, data + N);  
    int num = unique(data, data + N) - data;  
    for (int i = 0; i < N; ++i)  
        line[i].y2 = lower_bound(data, data + num, line[i].y2) - data + 1;  
    ll ans = 0;  
    int last = 0;  
    for (int i = 0; i < N; ++i)  
    {  
        if (i > 0 && line[i].y1 == line[i - 1].y1)  
        {  
            //++ans;  
            ans += i - last;  
        }  
        else  
        {  
            ans += i - query(line[i].y2 - 1);  
            last = i;  
        }  
        update(line[i].y2, 1);  
    }  
    printf("%lld\n", ans);  
    return 0;  
}