hdu 5480(维护前缀和+思路题)

Conturbatio

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 786 Accepted Submission(s): 358


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
For every query output "Yes" or "No" as mentioned above.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

Yes
No
Yes

Hint

Huge input, scanf recommended.

[align=left]Source[/align]
BestCoder Round #57 (div.2)

题意:在一个棋盘上有一些"车"，他能够攻击到与它同一行或者同一列的棋盘上的所有的格子，现在给出K个棋子的坐标，然后有Q组询问，每一次询问(x1,y1,x2,y2)这个方格内的所有棋子是否能够全部被攻击到。
题解:维护前缀和,统计 (x1-x2) 这一段区间里面的被攻击到的行的数量，统计(y1-y2)这一段区间里面的被攻击到的列的数量,如果sum(x1~x2) == x2-x1+1 ，那么这段区间全部能够被攻击到,列也就不用考虑了，对列的考虑亦如此。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;
const int N = 100005;
int flag_x
,flag_y
;
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
int n,m,k,q;
scanf("%d%d%d%d",&n,&m,&k,&q);
memset(flag_x,0,sizeof(flag_x));
memset(flag_y,0,sizeof(flag_y));
int x,y;
for(int i=1;i<=k;i++){
scanf("%d%d",&x,&y);
flag_x[x] = 1;
flag_y[y] = 1;
}
for(int i=1;i<=n;i++){
flag_x[i]+= flag_x[i-1];
}
for(int i=1;i<=m;i++){
flag_y[i] += flag_y[i-1];
}
while(q--){
int x1,y1,x2,y2;
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
if(flag_x[x2]-flag_x[x1-1]==x2-x1+1||flag_y[y2]-flag_y[y1-1]==y2-y1+1) printf("Yes\n");
else printf("No\n");
}
}
return 0;
}