poj 1050 To the Max（最大子矩阵和）

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 36079		Accepted: 18928

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：给出一个矩阵，求最大的子矩阵和。

思路：枚举每个子矩阵，然后压缩子矩阵（将每行同列的元素加起来，变成只有一行），使之变成一维的最大子序列和问题。

AC代码：

#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <vector>
#include <cstdlib>

using namespace std;
const double INF=100000000;
int max3(int a,int b,int c)
{
    if(a<b) a=b;
    return a>c?a:c;
}
int main()
{
    int n;
    int map[105][105],temp[105];
    while(cin>>n)
    {
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                cin>>map[i][j];
            int max=-INF;
        for(int i=1; i<=n; i++)
        {
            memset(temp,0,sizeof(temp));
            for(int j=i; j<=n; j++)
            {
                int sum=0;
                for(int k=1;k<=n;k++)       //压缩矩阵
                 temp[k]+=map[j][k];
                for(int k=1;k<=n;k++)    //求最大子序列和
                {
                    if(sum>=0)
                    sum+=temp[k];
                    else
                    sum=temp[k];
                    if(sum>max) max=sum;
                }
            }
        }
            cout<<max<<endl;
        }
        return 0;
    }