Maximum Sum
Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. 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. A sub-rectangle is any contiguous sub-array of size 1 × 1 or greater located within the whole array.
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-hand corner and has the 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 white-space (newlines and spaces). These N 2 integers make up the array in row-major order (i.e., all numbers on the first row, left-to-right, then all numbers on 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
The output is the sum of the maximal sub-rectangle.
Sample Input
input output
4
0 -2 -7 0
9 2 -6 2
-4 1 -4 1
-1 8 0 -2
15
最大子矩阵，不说了～
/*************************************************************************
> File Name: k.cpp
> Author:yuan
> Mail:
> Created Time: 2014年11月09日 星期日 22时38分42秒
************************************************************************/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
int mat[107][107];
int sum[107],dp[107];
int n,ans;
int main()
{
while(~scanf("%d",&n)){
memset(mat,0,sizeof(mat));
memset(sum,0,sizeof(sum));
ans=-0x3ffffff;
int m=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
scanf("%d",&mat[i][j]);
}
for(int i=1;i<=n;i++)
for(int j=i;j<=n;j++)
{
memset(sum,0,sizeof(sum));
for(int k=i;k<=j;k++)
{
for(int l=1;l<=n;l++)
sum[l]+=mat[k][l];
}
memset(dp,0,sizeof(dp));
for(int k=1;k<=n;k++)
{
if(dp[k-1]<0) dp[k]=sum[k];
else dp[k]=dp[k-1]+sum[k];
ans=max(ans,dp[k]);
}
}
printf("%d
",ans);
}
return 0;
}