poj 1050 To the Max（最大子矩阵和）

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal 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

40 -2 -7 0 9 2 -6 2-4 1 -4  1 -18  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;    }