hdu 5480 Conturbatio 线段树 单点更新，区间查询最小值

Conturbatio

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5480

Description

There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.

There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?

Input

The first line of the input is a integer T, meaning that there are T test cases.

Every test cases begin with four integers n,m,K,Q.
K is the number of Rook, Q is the number of queries.

Then K lines follow, each contain two integers x,y describing the coordinate of Rook.

Then Q lines follow, each contain four integers x1,y1,x2,y2 describing the left-down and right-up coordinates of query.

1≤n,m,K,Q≤100,000.

1≤x≤n,1≤y≤m.

1≤x1≤x2≤n,1≤y1≤y2≤m.

Output

For every query output "Yes" or "No" as mentioned above.

Sample Input

2
2 2 1 2
1 1
1 1 1 2
2 1 2 2
2 2 2 1
1 1
1 2
2 1 2 2

Sample Output

Yes
No
Yes

HINT

题意

在一个n \times mn×m的国际象棋棋盘上有很多车(Rook)，其中车可以攻击他所属的一行或一列，包括它自己所在的位置。
现在还有很多询问，每次询问给定一个棋盘内部的矩形，问矩形内部的所有格子是否都被车攻击到？

题解：

我是线段树做的，只要统计x1,x2这个区域的最小值和y1 y2这个区域的最小值都不同时为0 就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 150001
#define mod 10007
#define eps 1e-9
//const int inf=0x7fffffff;   //无限大
const int inf=0x3f3f3f3f;
/*
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
*/
//**************************************************************************************
 
struct data{
   int l,r,mn;
}tr[maxn*];
 
data tr2[maxn*];
void build2(int k,int s,int t)
{
    tr2[k].l=s;tr2[k].r=t;
    if(s==t){tr2[k].mn=;return;}
    int mid=(s+t)>>;
    build2(k<<,s,mid);
    build2(k<<|,mid+,t);
    tr2[k].mn=min(tr2[k<<].mn,tr2[k<<|].mn);
}
int ask2(int k,int s,int t)
{
    int l=tr2[k].l,r=tr2[k].r;
    if(s==l&&t==r)return tr2[k].mn;
    int mid=(l+r)>>;
    if(t<=mid)return ask2(k<<,s,t);
    if(s>mid)return ask2(k<<|,s,t);
    return min(ask2(k<<,s,mid),ask2(k<<|,mid+,t));
}
void update2(int k,int x,int y)
{
    int l=tr2[k].l,r=tr2[k].r;
    if(l==r){tr2[k].mn=y;return;}
    int mid=(l+r)>>;
    if(x<=mid)update2(k<<,x,y);
    if(x>mid)update2(k<<|,x,y);
    tr2[k].mn=min(tr2[k<<].mn,tr2[k<<|].mn);
}
void build(int k,int s,int t)
{
    tr[k].l=s;tr[k].r=t;
    if(s==t){tr[k].mn=;return;}
    int mid=(s+t)>>;
    build(k<<,s,mid);
    build(k<<|,mid+,t);
    tr[k].mn=min(tr[k<<].mn,tr[k<<|].mn);
}
int ask(int k,int s,int t)
{
    int l=tr[k].l,r=tr[k].r;
    if(s==l&&t==r)return tr[k].mn;
    int mid=(l+r)>>;
    if(t<=mid)return ask(k<<,s,t);
    if(s>mid)return ask(k<<|,s,t);
    return min(ask(k<<,s,mid),ask(k<<|,mid+,t));
}
void update(int k,int x,int y)
{
    int l=tr[k].l,r=tr[k].r;
    if(l==r){tr[k].mn=y;return;}
    int mid=(l+r)>>;
    if(x<=mid)update(k<<,x,y);
    if(x>mid)update(k<<|,x,y);
    tr[k].mn=min(tr[k<<].mn,tr[k<<|].mn);
}
 
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        int n,m,k,q;
        scanf("%d%d%d%d",&n,&m,&k,&q);
        build(,,n);
        build2(,,m);
        for(int i=;i<=k;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            update(,x,);
            update2(,y,);
        }
        for(int i=;i<=q;i++)
        {
            int x1,x2,y1,y2;
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            if(x1>x2)swap(x1,x2);
            if(y1>y2)swap(y1,y2);
            int d1 = max(ask(,x1,x2),ask2(,y1,y2));
            if(d1==)
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
    return ;
}