hdu 5480|| bestcoder ＃５７ div 2 Conturbatio（前缀和｜｜树状数组）

Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 211    Accepted Submission(s): 99

Problem 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

Huge input, scanf recommended.

 

Source

BestCoder Round #57 (div.2)

 

比较水．

唯一一点需要注意的是...

可能有重复元素...

因为我的思路是用两棵一维树状数组搞..

每个点标记为1

然后看矩形的两个方向中是否至少有一个方向上和等于长度...

所以这样如果有重复元素的话，不处理会出错..　

 

但实际上又没修改..直接前缀和就好了．．．

树状数组个毛线．．．

不过看到还有人线段树搞得233333

 

 

/*************************************************************************
    > File Name: code/bc/#57/1002.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年09月26日 星期六 19时31分10秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
#define lr dying111qqz
using namespace std;
#define For(i, n) for (int i=0;i<int(n);++i)  
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=1E5+7;
int c[N],d[N];
int n,m,K,Q;
bool vx[N],vy[N];

int lowbit(int x)
{
    return x&(-x);
}

void update ( int x,int delta)
{
    for ( int i = x ; i  <= n ; i = i + lowbit(i))
    {
    c[i] = c[i] + delta;
    }
}

LL sum ( int x)
{
    LL res = 0 ;
    for ( int i = x ; i >= 1 ; i = i - lowbit(i))
    {
     res  = res  + c[i];
    }
    return res;
}

void update2( int y,int delta)
{
    for ( int i = y ; i <= m ; i = i + lowbit(i))
    {
    d[i] = d[i] + delta;
    }
}

LL sum2( int y)
{
    LL res = 0 ;
    for ( int i = y  ;  i >= 1 ; i = i - lowbit(i))
    {
    res = res + d[i];
    }
    return res;

}
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif
   int T;
       scanf("%d",&T);
   while (T--)
    {
    ms(c,0);
    ms(d,0);
    memset(vx,false,sizeof(vx));
    memset(vy,false,sizeof(vy));
    scanf("%d %d %d %d",&n,&m,&K,&Q);
    for ( int i  = 0 ; i <K ; i++)
    {
        int x,y;
        scanf("%d %d",&x,&y);
        if (!vx[x])
        {
             update (x,1);
        vx [x] = true;
        }
        if (!vy[y])
        {
        update2 (y,1);
        vy[y] = true;
        }
        
    }
    for ( int i = 0 ; i < Q ; i++)
    {
        int x1,y1,x2,y2;
        scanf("%d %d %d %d",&x1,&y1,&x2,&y2);
        int xx = sum(x2)-sum(x1-1);
        int yy = sum2(y2)-sum2(y1-1);
      //  cout<<"sum(x2):"<<sum(x2)<<endl;
      //  cout<<"sum(x1-1):"<<sum(x1-1)<<endl;
      //  cout<<"sum(y2):"<<sum(y2)<<endl;
      //  cout<<"sum(y1-1)"<<sum(y1-1)<<endl;
      //  cout<<"xx:"<<xx<<endl;
      //  cout<<"yy:"<<yy<<endl;
        if (xx>=x2-x1+1||yy>=y2-y1+1)
        {
        puts("Yes");
        }
        else
        {
        puts("No");
        }
         
    }
    }
  
   
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}