最大子矩阵和

Problem Description

Given a two-dimensional array of positive and negative integers, a sub-rectangle is any contiguous sub-array of size 1 x 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.<br><br>As an example, the maximal sub-rectangle of the array:<br><br>0 -2 -7 0<br>9 2 -6 2<br>-4 1 -4 1<br>-1 8 0 -2<br><br>is in the lower left corner:<br><br>9 2<br>-4 1<br>-1 8<br><br>and has a sum of 15.<br>

 

Input

The input consists of an N x 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].<br>

 

Output

Output the sum of the maximal sub-rectangle.<br>

 

Sample Input

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

 

Sample Output

15

 

问题分析：（因为没有多组输入 WA。。。）

每次枚举子矩阵的最上行u和最下行d，再把这个子矩阵每一列的值相加（压缩成一行），压缩成一个一维数组，对于这个数组求其最大连续字段和，这样就相当于把把所有最上的行为u并且最下的行为d的最大子矩阵和求出来了。

AC代码：

#include <bits/stdc++.h>#define N 135using namespace std;int fun(int b[N],int n){    int i,max,c;    c=0,max=-100000;    for(i=1;i<=n;i++)    {        if(c>0)            c=c+b[i];        else            c=b[i];        if(max<c)            max=c;    }    return max;}int main(){    ios::sync_with_stdio(false);    int i,j,n,max,sum,k;    int a[N][N],b[N];  while(cin>>n)    {    for(i=1;i<=n;i++)        for(j=1;j<=n;j++)        cin>>a[i][j];    max=-100000;    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)            b[j]=0;            for(j=i;j<=n;j++)            {                for(k=1;k<=n;k++)                    b[k]+=a[j][k];                sum=fun(b,n);                if(max<sum)                    max=sum;            }    }    cout<<max<<endl;    }    return 0;}