POJ 1050 To the Max

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

给你一个矩阵，让你求最大子矩阵和。因为n只有100，所以可以用n^3的方法解决，就是可以把两行或者多行合并成一行，然后就像求最大子串和一样去求出最大值就好了。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define LL long longusing namespace std;int n;int a[110][110],b[110];int fun(){    int maxn = 0;    int sum = 0;    for(int i=1;i<=n;i++)    {        if(sum > 0)            sum += b[i];        else            sum = b[i];        maxn = max(sum,maxn);    }    return maxn;}int main(void){    int i,j,k;    while(scanf("%d",&n)==1)    {        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)                scanf("%d",&a[i][j]);        int maxn = 0;        for(i=1;i<=n;i++)        {            memset(b,0,sizeof(b));            for(j=i;j<=n;j++)            {                for(k=1;k<=n;k++)                    b[k] += a[j][k];                maxn = max(maxn,fun());            }        }        printf("%d\n",maxn);    }    return 0;}