Maximum Sum

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

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 N² integers separated by white-space (newlines and spaces). These N² 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:

　　　因为所求之和为矩阵和，所以必须满足行数相邻，每行起点终点相同，即需要构造dp[i][j][k],表示矩阵行数以终点为i，以j为起点，k为终点的矩阵最大和，sum值记录该i行起点为for(k--),终点为k的矩阵和。当然同样可以枚举以k为起点的矩阵和，同理。

　　　复杂度O(n3);

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
const int INF = 0x3f3f3f3f;
using namespace std;
int a[105][105];
int dp[105][105][105];
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++){
        for(int j = 1; j <= n; j++){
            int sum = 0;
            for(int k = j; k >= 1; k--)//枚举以k为终点;//（for(int k = j; k <= n; j++)//这是枚举k为起点;）
　　　　　　　 {
                sum += a[i][k];//sum值记录第i行以k为终点的矩阵和;
                dp[i][j][k] = max(sum,sum+ dp[i-1][j][k]);
                ans = max(ans, dp[i][j][k]);
            }
        }
    }
    printf("%d", ans);
    return 0;
}