题目
Check Corners
Time Limit: 2000/10000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1450 Accepted Submission(s): 534
Problem Description
to know the number of them. But soon he find that it is too complex, so he changes his mind, he just want to know whether there is a maximum at the four corners of the sub-matrix, he calls this “Check corners”. It’s a boring job when selecting too many sub-matrices,
so he asks you for help. (For the “Check corners” part: If the sub-matrix has only one row or column just check the two endpoints. If the sub-matrix has only one entry just output “yes”.)
Input
For each test case, the first line contains two integers m, n (1 <= m, n <= 300), which is the size of the row and column of the matrix, respectively. The next m lines with n integers each gives the elements of the matrix which fit in non-negative 32-bit integer.
The next line contains a single integer Q (1 <= Q <= 1,000,000), the number of queries. The next Q lines give one query on each line, with four integers r1, c1, r2, c2 (1 <= r1 <= r2 <= m, 1 <= c1 <= c2 <= n), which are the indices of the upper-left corner
and lower-right corner of the sub-matrix in question.
Output
Sample Input
4 4 4 4 10 7 2 13 9 11 5 7 8 20 13 20 8 2 4 1 1 4 4 1 1 3 3 1 3 3 4 1 1 1 1
Sample Output
20 no 13 no 20 yes 4 yes
Source
Recommend
题解
和前一发文章是类似的，只不过这个是最大值，上一发是最小值。
那我就不多说了，直接上代码了。
代码示例
#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
/*
* 二维RMQ，预处理复杂度 n*m*log*(n)*log(m)
* 数组下标从1开始
*/
int val[310][310];
int dp[310][310][9][9];//最大值
int mm[310];//二进制位数减一，使用前初始化
void initRMQ(int n,int m){
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
dp[i][j][0][0] = val[i][j];
for(int ii = 0; ii <= mm[n]; ii++)
for(int jj = 0; jj <= mm[m]; jj++)
if(ii+jj)
for(int i = 1; i + (1<<ii) - 1 <= n; i++)
for(int j = 1; j + (1<<jj) - 1 <= m; j++)
{
if(ii)
dp[i][j][ii][jj] = max(dp[i][j][ii-1][jj],dp[i+(1<<(ii-1))][j][ii-1][jj]);
else
dp[i][j][ii][jj] = max(dp[i][j][ii][jj-1],dp[i][j+(1<<(jj-1))][ii][jj-1]);
}
}
//查询矩形内的最大值(x1<=x2,y1<=y2)
int rmq(int x1,int y1,int x2,int y2){
int k1 = mm[x2-x1+1];
int k2 = mm[y2-y1+1];
x2 = x2 - (1<<k1) + 1;
y2 = y2 - (1<<k2) + 1;
return max(max(dp[x1][y1][k1][k2],dp[x1][y2][k1][k2]),max(dp[x2][y1][k1][k2],dp[x2][y2][k1][k2]));
}
int main(){
//在外面对mm数组进行初始化
mm[0] = -1;
for(int i = 1; i <= 305; i++)
mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
int n,m;
int Q;
int r1,c1,r2,c2;
while(scanf("%d%d",&n,&m) == 2){
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
scanf("%d",&val[i][j]);
initRMQ(n,m);
scanf("%d",&Q);
while(Q--){
scanf("%d%d%d%d",&r1,&c1,&r2,&c2);
if(r1 > r2)swap(r1,r2);
if(c1 > c2)swap(c1,c2);
int tmp = rmq(r1,c1,r2,c2);
printf("%d ",tmp);
if(tmp == val[r1][c1] || tmp == val[r1][c2] || tmp == val[r2][c1] || tmp == val[r2][c2])
printf("yes\n");
else printf("no\n");
}
}
return 0;
}