hdu1081 To the Max

To the Max

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 20000/10000K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 1
Problem 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
4
0 -2 -7 0 9 2 -6 2
-4 1 -4 1 -1

8 0 -2

Sample Output
15

求最大矩阵和

做dp[i][j]表示前i行前j列最大值
则有dp[i][j]=max(dp[i][j],dp[i][j-1]+a[i][j])

#include <cstdio>#include <algorithm>#include <cstring>#include <iostream>#define INF 0xffffusing namespace std;int main(){    int n;    int a[150][150];    int dp[150][150];///dp[i][j]表示在i行时前j列的最大值    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        for(int i=1; i<=n; i++)            for(int j=1; j<=n; j++)                {                    scanf("%d",&a[i][j]);                    dp[i][j]+=dp[i][j-1]+a[i][j];                }        //dp[0][0]=a[0][0];        int maxn=-100000;        for(int i=1;i<=n;i++)        {            //dp[i][0]=max(dp[i][0],dp[i-1][0]+a[i][j]);            for(int j=i;j<=n;j++)            {                int sum=0;                for(int k=1;k<=n;k++)                {                    if(sum<0)                    sum=0;                    sum+=dp[k][j]-dp[k][i-1];                    if(sum>maxn)                    maxn=sum;                }                //dp[i][j]=max(dp[i][j],dp[i-1][j]+dp[i][j-1]+a[i-1][j]+a[i][j-1]);            }        }        cout<<maxn<<endl;    }    return 0;}