简单DP+暴力 POJ 1050

To the Max

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 45915		Accepted: 24282

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

 /*这道题目n^3居然不会超时，暴力就好了*/
 #include<iostream>
 using namespace std;
 #include<cstdio>
 #define N 101
 int sum[N][N]={0},n,x;
 int main()
 {
     scanf("%d",&n);
     for(int i=1;i<=n;++i)
       for(int j=1;j<=n;++j)
       {
            scanf("%d",&x);
            sum[i][j]=sum[i][j-1]+x;/*sum[i][j]表示的是第i行前j个数的和*/
       }
     int ans=-(1<<30);
     for(int i=1;i<=n;++i)
       for(int j=i;j<=n;++j)/*暴力枚举每行区间*/
       {
           int tmp=0;/*tmp为当前矩阵的大小*/
           for(int k=1;k<=n;++k)/*枚举每一行*/
           {
               int que=sum[k][j]-sum[k][i-1];/*取出这一行*/
               if(tmp>0) tmp+=que;/*如果当前矩阵的大小已经<0了，那么再加上就要放弃之前的矩阵，放弃一定会更优*/
               else tmp=que;
               ans=max(ans,tmp);/*因为我们会随时放弃矩阵，所以最大值的更新，在循环中进行*/
           }
       }
     cout<<ans;
     return 0;
 }