To The Max(dp思想)

http://acm.hdu.edu.cn/showproblem.php?pid=1081

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10135    Accepted Submission(s): 4885

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.

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 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].

 

 

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

 

思路还是不够

//dp有一个特点就是要求出结果必须枚举所有情况，这根搜索有很大的区别#include <cstdio>int dp[110][110];const int maxn=1<<30;int main(){    int n;    while(scanf("%d",&n)!=EOF){        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++){                int a;                scanf("%d",&a);                dp[i][j]=dp[i][j-1]+a;//dp思想,再次想到了记忆化搜索            }        int ans=-maxn;        for(int i=1;i<=n;i++)            for(int j=i;j<=n;j++){//ij分别表示子矩阵的某一行开头和结尾，这里少掉一维的话就不好表示了                int sum=0;                for(int k=1;k<=n;k++){//这里是一维求最大连续和的思想                    if(sum<0)sum=0;                    sum+=dp[k][j]-dp[k][i-1];                    if(sum>ans)ans=sum;                }            }//枚举了所有的情况    printf("%d\n",ans);    }    return 0;}