HDU 5480（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): 457    Accepted Submission(s): 208

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

22 2 1 21 11 1 1 22 1 2 22 2 2 11 11 22 1 2 2

 

Sample Output

YesNoYes
Hint
Huge input, scanf recommended.
 

 

//开两个数组将行和列  出现的个数累加  询问时只要行或者列个数之和等于x2-x1+1||y2-y1+1

#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#include <map>#include <queue>using namespace std;const int maxn=100000+10;const int inf=(1<<30);int row[maxn],col[maxn];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m,k,q;        memset(row,0,sizeof(row));        memset(col,0,sizeof(col));        scanf("%d%d%d%d",&n,&m,&k,&q);        while(k--)        {            int a,b;            scanf("%d%d",&a,&b);            row[a]=1,col[b]=1;  //赋值为1  有可能出现相同的不能累加        }        for(int i=1;i<=n;i++)        {            row[i]+=row[i-1];        }        for(int i=1;i<=m;i++)        {            col[i]+=col[i-1];        }        while(q--)        {            int x1,y1,x2,y2,res1,res2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            res1=row[x2]-row[x1-1];            res2=col[y2]-col[y1-1];            if(res1==(x2-x1+1)||res2==(y2-y1+1))                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}

树状数组 -- 思路一样

#include <iostream>#include <stdio.h>#include <string.h>#include <cmath>#include <algorithm>#include <map>#include <queue>using namespace std;const int maxn=100000+10;const int inf=(1<<30);int row[maxn],col[maxn];bool vis[maxn],vis1[maxn];int lowbit(int x){    return x&(-x);}void add(int i,int val,int c[],int k){    while(i<=k)    {        c[i]+=val;        i+=lowbit(i);    }}int sum(int i,int c[]){    int sum=0;    while(i)    {        sum+=c[i];        i-=lowbit(i);    }    return sum;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m,k,q;        memset(row,0,sizeof(row));        memset(col,0,sizeof(col));        memset(vis,0,sizeof(vis));        memset(vis1,0,sizeof(vis1));        scanf("%d%d%d%d",&n,&m,&k,&q);        while(k--)        {            int a,b;            scanf("%d%d",&a,&b);            if(!vis[a])   //判断是否出现过                add(a,1,row,n);            if(!vis1[b])                add(b,1,col,m);            vis[a]=vis1[b]=1;        }        while(q--)        {            int x1,y1,x2,y2,res1,res2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            res1=sum(x2,row)-sum(x1-1,row);            res2=sum(y2,col)-sum(y1-1,col);            if(res1==(x2-x1+1)||res2==(y2-y1+1))                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}