Matrix - HDU 2686 dp

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本文探讨了在矩阵中从左上角到右下角再返回左上角的最大路径求和问题，通过动态规划方法解决，并提供了实现代码。

Matrix

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1548 Accepted Submission(s): 834

Problem Description

Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.
Every time yifenfei should to do is that choose a detour which frome the top left point to the bottom right point and than back to the top left point with the maximal values of sum integers that area of Matrix yifenfei choose. But from the top to the bottom can only choose right and down, from the bottom to the top can only choose left and up. And yifenfei can not pass the same area of the Matrix except the start and end.

Input

The input contains multiple test cases.
Each case first line given the integer n (2<n<30)
Than n lines,each line include n positive integers.(<100)

Output

For each test case output the maximal values yifenfei can get.

Sample Input

Sample Output

28
46
80

题意：从左上角到右下角再返回左上角，不能走到重复的格子，问你最大的路径的权值和是多少。

思路：先循环要走多少步，每步都对应有右边走的点和下边走的点可能落在哪些地方，对于这些情况，考虑他的上一步可以走到这种情况的情况，然后dp。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[35][35][35][35],value[35][35];
int main()
{ int n,i,j,k,x1,x2,y1,y2;
  while(~scanf("%d",&n))
  { for(i=1;i<=n;i++)
     for(j=1;j<=n;j++)
      scanf("%d",&value[i][j]);
    memset(dp,0,sizeof(dp));
    for(k=1;k<=2*n-3;k++)
     for(x1=2;x1<=1+k && x1<=n;x1++)
     { y1=2+k-x1;
       for(y2=2;y2<=1+k && y2<=n;y2++)
       { x2=2+k-y2;
         if(x1>x2 && y2>y1)
         { dp[y1][x1][y2][x2]=max(dp[y1][x1-1][y2][x2-1],dp[y1-1][x1][y2-1][x2]);
           dp[y1][x1][y2][x2]=max(dp[y1][x1][y2][x2],dp[y1-1][x1][y2][x2-1]);
           if(!(x1-x2==1 && y2-y1==1))
            dp[y1][x1][y2][x2]=max(dp[y1][x1][y2][x2],dp[y1][x1-1][y2-1][x2]);
           dp[y1][x1][y2][x2]+=value[y1][x1]+value[y2][x2];
         }
       }
     }
    printf("%d\n",dp[n-1][n][n][n-1]+value[1][1]+value[n][n]);
  }
}