hdu 1081 To the Max

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1081

题目描述：

Description

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]. 

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

题目大意：

在一个n*n的矩阵中，找到一个子矩阵，使得和最大

题目分析：

同找最大和子序列，只是把二维变一维，用到了状态压缩，这里我把列压缩

AC代码：

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int dp[105][105];int main(){    int n,num;    while(scanf("%d",&n)!=EOF)    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                scanf("%d",&num);                dp[i][j]=dp[i][j-1]+num;//代表第i行第一个数字到第j个数字的和            }        }        int Max=-0xfffffff,sum=0;        for(int i=1;i<=n;i++)        {            for(int j=1;j<i;j++)//i j 之间的列数的和当做整体            {                sum=0;                for(int k=1;k<=n;k++)//k行                {                    if(sum<0)                        {                            sum=0;                        }                    sum+=(dp[k][i]-dp[k][j-1]);                    if(sum>Max)                        {                            Max=sum;                        }                }            }        }        cout<<Max<<endl;    }    return 0;}