hdu 1081To The Max(二维最大连续子串和)

To The Max

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6994    Accepted Submission(s): 3375

Problem 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：给你一个n*n的矩阵，让你找一个子矩阵，使得这个矩阵内所有数字的和最大。

    解题思路：先联系一维的最大连续子串和，dp[i]=max(dp[i-1],0)+a[i]其中dp[i]表示前面i项内所取得的最大值，必须包含第i项。a[i]表示第i项。这个二维矩阵的题目也可以进行转换。先用mp[i][j]，表示第i行前j列的和。那么状态转移方程可以转换为dp[k]=max(dp[k-1],0)+a[k]，我们可以将a[k]看作第j列到第i列的第k项的和，a[k]=mp[k][i]-mp[k][j-1].

     题目地址：To The Max

AC代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int mp[105][105];int main(){    int i,j,k,res,n,x;    while(cin>>n)    {        res=-1e9;        memset(mp,0,sizeof(mp));        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                scanf("%d",&x);                mp[i][j]=mp[i][j-1]+x;            }        for(i=1;i<=n;i++)        {            for(j=1;j<=i;j++)    //从第j列到第i列            {                int tmp=-1;                for(k=1;k<=n;k++)   //看成一维的 dp[k]=max(dp[k-1],0)+a[k]                {                    tmp=max(tmp,0)+mp[k][i]-mp[k][j-1];                    res=max(res,tmp);                }            }        }        cout<<res<<endl;    }    return 0;}