Hdu 5480 Conturbatio

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Conturbatio

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 341    Accepted Submission(s): 155


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.

1n,m,K,Q100,000.

1xn,1ym.

1x1x2n,1y1y2m.
 

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.
 

Source
BestCoder Round #57 (div.2)
 

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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:记录下行攻击的前缀和及列攻击的前缀和,每个矩阵只要满足行攻击和列攻击全部攻击到就是yes.

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <string>#include <map>#include <cstdlib>#include <cmath>#include <vector>#include <set>#include <queue>using namespace std;typedef long long ll;const int maxn=1e5+10;int n,m,k,q;int cs[maxn],rs[maxn];void init() {    memset(cs,0,sizeof cs);//hang    memset(rs,0,sizeof rs);//lie}int main() {    int t;    cin>>t;    int ca=1;    while(t--) {        scanf("%d%d%d%d",&n,&m,&k,&q);        init();        for(int i=0; i<k; i++) {            int x,y;            scanf("%d%d",&x,&y);            cs[x]=1;            rs[y]=1;        }        for(int i=1; i<=n; i++)            cs[i]+=cs[i-1];        for(int i=1; i<=m; i++)            rs[i]+=rs[i-1];        while(q--) {            int lx,ly,rx,ry;            scanf("%d%d%d%d",&lx,&ly,&rx,&ry);            if(cs[rx]-cs[lx-1]==rx-lx+1||(rs[ry]-rs[ly-1]==ry-ly+1)) {                printf("Yes\n");            } else {                printf("No\n");            }        }    }    return 0;}


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