poj1050（DP）

http://poj.org/problem?id=1050

To the Max

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 31034Accepted: 16113

Description

Given atwo-dimensional array of positive and negative integers, asub-rectangle is any contiguous sub-array of size 1*1 or greaterlocated within the whole array. The sum of a rectangle is the sumof all the elements in that rectangle. In this problem thesub-rectangle with the largest sum is referred to as the maximalsub-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 consistsof an N * N array of integers. The input begins with a singlepositive integer N on a line by itself, indicating the size of thesquare two-dimensional array. This is followed by N^2 integersseparated by whitespace (spaces and newlines). These are the N^2integers of the array, presented in row-major order. That is, allnumbers in the first row, left to right, then all numbers in thesecond row, left to right, etc. N may be as large as 100. Thenumbers in the array will be in the range [-127,127].

Output

Output the sum ofthe maximal sub-rectangle.

Sample Input

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

Sample Output

15

Source

最大子矩阵可以参考http://www.sunhongfeng.com/2010/04/poj1050-to-the-max/或者

http://www.cnblogs.com/aiyite826/archive/2010/07/23/1784032.html

#include<stdio.h>
#include<string.h>
int m[105][105],dp[105],t[105];
int main()
{
 int n;
 while(scanf("%d",&n)!=EOF)
 {
  int i,j,k;
  for(i=1;i<=n;i++)
   for(j=1;j<=n;j++)
    scanf("%d",&m[i][j]);
  int max=0;
  for(i=1;i<=n;i++)//i表示从第几行开始
  {
   memset(t,0,sizeof(t));
   for(j=i;j<=n;j++)//j表示从第几行结束
   {
    for(k=1;k<=n;k++)//k表示第几列
    {
     t[k]+=m[j][k];
     if(t[k]+dp[k-1]>0)//开始算一维数组的最大子串
      dp[k]=t[k]+dp[k-1];
     else
      dp[k]=0;
     if(max<dp[k])
      max=dp[k];
    }
   }
  }
  printf("%d\n",max);
 }
 return 0;
}