poj 动态规划之1050 To the Max
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poj 1050 To the Max
Description
Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1*1 or greater located within the whole array. The sum of a rectangle is the sum of all the elements in that rectangle. In this problem the sub-rectangle with the largest sum is referred to as the maximal sub-rectangle.
As an example, the maximal sub-rectangle of the array:
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
is in the lower left corner:
9 2
-4 1
-1 8
and has a sum of 15.
Input
The input consists of an N * N array of integers. The input begins with a single positive integer N on a line by itself, indicating the size of the square two-dimensional array. This is followed by N^2 integers separated by whitespace (spaces and newlines). These are the N^2 integers of the array, presented in row-major order. That is, all numbers in the first row, left to right, then all numbers in the second row, left to right, etc. N may be as large as 100. The numbers in the array will be in the range [-127,127].
Output
Output the sum of the maximal sub-rectangle.
Sample Input
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1
8 0 -2
Sample Output
15
解释:为了避免计算量,设置sum[i][j]代表第i行1到j列的和,其余的就是正常的dp
#include<cstdio>#include<string>#include<iostream>#include<algorithm>#include<memory.h>using namespace std;int sum[105][105], a[105][105], dp[105];int main(){ int N, i, j, k, ans; ans = 0; scanf("%d", &N); memset(sum, 0, sizeof(sum)); memset(dp, 0, sizeof(dp)); for (i = 1; i <= N; i++) for (j = 1; j <= N; j++) { scanf("%d", &a[i][j]); sum[i][j] = sum[i][j - 1] + a[i][j]; } for (i = 1; i <= N;i++) for (j = i; j <= N; j++) { dp[1] = sum[1][j] - sum[1][i - 1]; for (k = 2; k <= N; k++) { dp[k] = sum[k][j] - sum[k][i - 1]; if (dp[k - 1] > 0) dp[k] += dp[k - 1]; ans = max(ans, dp[k]); } } printf("%d", ans); return 0;}
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