POJ 1050：To the Max

To the Max

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 43241 Accepted: 22934

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

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

Sample Output

15

题意是给定一个矩阵，求其子矩阵的最大和。

这题也是弄得相当郁闷，一开始暴力，结果预料之中的TLE。然后试了一下dp，结果还MLE。。。郁闷得不行。

然后看了别人的思路，发现可以二维变一维，想了想忽然恍然大悟。

将每一列的加起来，就是一维了。枚举不同行即可。之前怎么做的这次怎么求。

代码：

#include <iostream>#include <string>#include <cstring>#include <algorithm>#include <cmath>#pragma warning(disable:4996) using namespace std;int value[250][250];int value2[250];int dp[250];int main(){//freopen("input.txt","r",stdin);//freopen("out.txt","w",stdout);int N,i,j,h,k,g,f;int ans=-100;scanf("%d",&N);memset(dp,0,sizeof(dp));memset(value2,0,sizeof(value2));for(i=1;i<=N;i++){for(j=1;j<=N;j++){scanf("%d",&value[i][j]);ans=max(ans,value[i][j]);}}for(i=1;i<=N;i++){for(h=i;h<=N;h++){for(k=1;k<=N;k++){value2[k] += value[h][k];dp[k] = max(dp[k-1]+value2[k],value2[k]);ans = max(ans,dp[k]);}memset(dp,0,sizeof(dp));}memset(value2,0,sizeof(value2));}cout<<ans<<endl;return 0;}