poj 1050 To the Max

DP+暴力

提交地址：http://poj.org/problem?id=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
提议大体上就是求n*n的矩阵中的最大矩形面积，不同与以往的情况就是里边得值有可能是负的（n^3不会超时哦）

#include<cstdio>
#include<cstring>
#include<iostream>

#define N 101

using namespace std;

int sum[N][N];
int n,x;
int num=-(1<<30);

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]表示第i行前j个数的和
          }
      for(int i=1;i<=n;i++)
        for(int j=i;j<=n;j++)                     //i、j表示的是举行的长度 
          {
              int t=0;
              for(int k=1;k<=n;k++)               //k表示举行的高度 
                {
                    int q=sum[k][j]-sum[k][i-1];    //枚举所有长为i—j的高为k的矩形，把他当前行的i—j位提取出来 
                                                    //再加上它之上几行的i—j的矩形的面积，就是一个新矩形的面积
                                                //本题和其他动规题一样，都是先处理小的，在由小的得到大的 
                    if(t>0) t+=q;                   //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    t=q;
                    num=max(num,t);                 //比较，取优 
                }
          }
    printf("%d",num);
      return 0;
  }