POJ 1050 To the Max - (DP)

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题目链接：http://poj.org/problem?id=1050
复杂度 $O(N^3)$，选取两列，然后把他们之间的矩阵压缩成一条竖线，
求一维连续最大字段和。

#include <stdio.h>  
#include <vector>   
#include <algorithm>
using namespace std;

//POJ 1050 To the Max

int matrix[101][101];
int sumrow[101][101];

int main()
{
    int N;
    scanf("%d", &N);
    const int l = N * N;
    for (int i = 1; i <= N; i++)
        for (int j = 1; j <= N; j++)
        {
            scanf("%d", &matrix[i][j]);
            sumrow[i][j] = sumrow[i][j - 1] + matrix[i][j];
        }
    int _max = 0x80000000;
    for (int start = 1; start <= N; start++)
        for (int end = start; end <= N; end++)
        {
            vector<int> dp(N + 1), v(N + 1);
            for (int row = 1; row <= N; row++)
                v[row] = sumrow[row][end] - sumrow[row][start - 1];
            dp[1] = v[1];
            _max = max(dp[1], _max);
            for (int i = 2; i <= N; i++)
            {
                if (dp[i - 1] > 0)
                    dp[i] = v[i] + dp[i - 1];
                else dp[i] = v[i];
                _max = max(_max, dp[i]);
            }
        }
    printf("%d", _max);
    system("pause");
    return 0;
}