POJ-1050To the Max

To the Max

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

总结
     虽然做了dp有一段时间了，但感觉还是一脸懵逼，看到这道题时完全没有思路，然后又是各种翻博客，看了半天才明白应该用什么思路，或许我就是一个
见得多才做的了题的人吧，以前弄奥数也是，很少有自己一次性做出来的新题，我要总结过这些思路后才想到怎么做。总之，在这个题我也学到了许多，
如何将二维转化为一维，如何求最长子串（请原谅我忘记了o(╯□╰)o）。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

#define inf 0x3f3f3f3f;

int mat[105][105];
int temp[105];
int n;

int main()
{
    int max;
    cin>>n;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&mat[i][j]);
        }
    }
    max=-inf;
    for(int i=1;i<n;i++){
        memset(temp,0,sizeof(temp));
        for(int j=i;j<=n;j++){
            int sum=0,max1=0;
            for(int k=1;k<=n;k++){
                temp[k]+=mat[j][k];
                sum+=temp[k];
                if(sum<0) sum=0;
                if(sum>max1) max1=sum;
            }
            if(max<max1) max=max1;
        }
    }
    cout <<max<<endl;
    return 0;
}