poj1050 To the Max（降维dp）

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 49351		Accepted: 26142

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

题意：

求矩阵中和最大的子矩阵。

思路：

把题目转化成一维的最大连续子段和。

枚举起始行i和终止行j，将i～j行的数对应相加，求解最大连续子段和。

代码：

#include<iostream>
using namespace std;
const int maxn = 105;
int grap[maxn][maxn], sum[maxn][maxn];
int main()
{
    int n, tmp, ans=-0x3f3f3f3f;
    cin>>n;
    for(int i=1; i<=n; ++i)
    {
        for(int j=1; j<=n; ++j)
        {
            cin>>grap[i][j];
            sum[i][j]=sum[i-1][j]+grap[i][j];
        }
    }
    for(int i=1; i<=n; ++i)
    {
        for(int j=i; j<=n; ++j)
        {
            tmp=0;
            for(int k=1; k<=n; ++k)
            {
                if(tmp<=0)
                    tmp=sum[j][k]-sum[i-1][k];
                else
                    tmp+=sum[j][k]-sum[i-1][k];
                ans=max(ans, tmp);
            }
        }
    }
    cout<<ans<<endl;
    return 0;
}