Hduoj1559【DP】

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/*最大子矩阵Time Limit: 30000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3280    Accepted Submission(s): 1652Problem Description给你一个m×n的整数矩阵，在上面找一个x×y的子矩阵，使子矩阵中所有元素的和最大。 Input输入数据的第一行为一个正整数T，表示有T组测试数据。每一组测试数据的第一行为四个正整数m,n,x,y（0<m,n<1000 AND 0<x<=m AND 0<y<=n），表示给定的矩形有m行n列。接下来这个矩阵，有m行，每行有n个不大于1000的正整数。 Output对于每组数据，输出一个整数，表示子矩阵的最大和。Sample Input14 5 2 23 361 649 676 588992 762 156 993 169662 34 638 89 543525 165 254 809 280Sample Output2474AuthorlwgSourceHDU 2006-12 Programming Contest RecommendLL   |   We have carefully selected several similar problems for you:  1505 1257 1024 1501 1494 */ #include<stdio.h>#include<string.h>int s[1001][1001];int main(){int i, j, t, n, m, x, y;scanf("%d", &t);while(t--){int max = 0;scanf("%d%d%d%d", &m, &n, &x, &y);for(i = 0; i < m; ++i){for(j = 0; j < n; ++j){int a;if(i == 0 && j == 0)scanf("%d", &s[i][j]);else{scanf("%d", &a);if(i == 0 && j != 0)s[i][j] = s[i][j-1] + a;else if( i != 0 && j == 0)s[i][j] = s[i-1][j] + a;elses[i][j] = s[i-1][j] + s[i][j-1] - s[i-1][j-1] + a;}}}for(i = 0; i + x <= m; ++i){for(j = 0; j + y <= n; ++j){if(i != 0 && j != 0){if(s[i+x-1][j+y-1] - s[i+x-1][j-1] - s[i-1][j+y-1] + s[i-1][j-1] > max)max = s[i-1+x][j-1+y] - s[i-1+x][j-1] - s[i-1][j-1+y] + s[i-1][j-1];}else if(i == 0 && j != 0){if(s[i+x-1][j+y-1] - s[i+x-1][j-1] > max)max = s[i+x-1][j+y-1] - s[i+x-1][j-1];}else if(i != 0 && j == 0){if(s[i+x-1][j+y-1] - s[i-1][j+y-1] > max)max = s[i+x-1][j+y-1] - s[i-1][j+y-1];}else{if(s[i+x-1][j+y-1] > max)max = s[i+x-1][j+y-1];}}}printf("%d\n", max);}return 0;}

题意：很明显。

思路：思路就是dp了，一边输入一边求起点到输入点的矩阵和，最后一个一个找符合矩阵（x，y）的最大值。

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