poj_1050To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 36015 Accepted: 18897

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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  0 -2

Sample Output

15

果将每一个二维矩阵的每一列的数据总和,按照列的序号分别对应地保存到一个一维数组中,就转化成一维,其性质跟求最大子段和一样,而在这里只不过要进行多次这样的计算而已 ^ -^ 经典Dynamic programming 中的经典!!!!

#include <iostream>#include <cstring>using namespace std;#pragma warning(disable : 4996)const int MAXN = 105;int arr[MAXN][MAXN], dp[MAXN];int n;int DP(){int thissum, maxsum;thissum = maxsum = 0;for (int i = 1; i <= n; i++){thissum += dp[i];if(thissum > maxsum){maxsum = thissum;}if(thissum < 0){thissum = 0;}}return maxsum;}int main(){freopen("in.txt", "r", stdin);int i, j, k;int sum, ans;while(scanf("%d", &n) != EOF){for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){scanf("%d", &arr[i][j]);}}ans = 0;for (int i = 1; i <= n; i++){memset(dp, 0, sizeof(dp));for (int j = i; j <= n; j++){for (k = 1; k <= n; k++){dp[k] += arr[j][k];}sum = DP();if(sum > ans){ans = sum;}}}printf("%d\n", ans);}}