DP(4)

Maximum Sum

Time Limit:500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice URAL 1146

Description

Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.

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-hand corner and has the 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 white-space (newlines and spaces). These N ² integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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

The output is the sum of the maximal sub-rectangle.

Sample Input

input	output
4 0 -2 -7 0 9 2 -6 2 -4 1 -4 1 -1 8 0 -2	15

DP

#include<cstdio>
#include<algorithm>
const int INF = 0x3f3f3f3f;
const int maxn = 105;
using namespace std;
int a[maxn][maxn];
int dp[maxn][maxn][maxn];
int main(){
    int n;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= n; j++)
            scanf("%d",&a[i][j]);
    int ans = -INF;
    for(int i = 1; i <= n; i++)//枚举每列
        for(int j = 1; j <= n; j++){//枚举此列的点
            int sum = 0;
            for(int k = j; k >= 1; k--){//枚举该点与其他点的构成的矩形
                sum += a[i][k];//该矩形总值
                dp[i][k][j] = max(sum,sum+dp[i-1][k][j]);//加上相邻列构成的矩形的总值，进行比较。
                ans = max(dp[i][k][j],ans);//得出max总值。
            }
    }
    printf("%d\n",ans);
}