hdu1081【模拟】

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 −127,127−127,127. 

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：78ms过题，应该是数据比较水吧，补一篇博文；

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define max_n 110using namespace std;typedef long long LL;int a[max_n][max_n],c[max_n][max_n];int main(){int n;while(~scanf("%d",&n)){memset(a,0,sizeof(a));for(int i=1;i<=n;i++)for(int j=1;j<=n;j++)scanf("%d",&c[i][j]);int ans=0,sum=0,maxn=-100001200;for(int i=1;i<=n;i++){sum=0;for(int j=1;j<=n;j++){for(int p=1;p<=i;p++){sum+=c[p][j];}a[i][j]=sum;}}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){for(int p=0;p<i;p++){for(int u=0;u<j;u++){ans=a[i][j]-a[p][j]-a[i][u]+a[p][u];//减的有重叠部分 maxn=max(maxn,ans);}}}}printf("%d\n",maxn);}return 0; }