【NJUST5480】Conturbatio
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Conturbatio
Time Limit: 6000/3000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
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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?
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.
$1\leq n , m , K , Q \leq 100,000$.
$1\leq x \leq n , 1 \leq y \leq m$.
$1\leq x1 \leq x2 \leq n , 1 \leq y1 \leq y2 \leq m$.
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.
$1\leq n , m , K , Q \leq 100,000$.
$1\leq x \leq n , 1 \leq y \leq m$.
$1\leq x1 \leq x2 \leq n , 1 \leq y1 \leq y2 \leq m$.
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
YesNoYesHintHuge input, scanf recommended.
Hint
hujie
Source
BestCoder Round #57 (div.2)
一道比较好的思维题目
#include<cstdio>#include<iostream>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int N = 100000+10;int row[N],col[N];int main() {int T;scanf("%d",&T);while(T--) {memset(row,0,sizeof(row));memset(col,0,sizeof(col));int n,m,k,q;scanf("%d%d%d%d",&n,&m,&k,&q);for(int i=0; i<k; i++) {int a,b;scanf("%d%d",&a,&b);row[a]=1;col[b]=1;}for(int i=1; i<=n; i++) {row[i]=row[i]+row[i-1];}for(int i=1; i<=m; i++) {col[i]=col[i]+col[i-1];}while(q--) {int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);if(x2-x1+1==row[x2]-row[x1-1]||y2-y1+1==col[y2]-col[y1-1])printf("Yes\n");elseprintf("No\n");}}return 0;}题目地址:http://acm.hust.edu.cn/vjudge/contest/127335#problem/A
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