Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 848    Accepted Submission(s): 484


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
2 10 3 5 10 3 10 3 3 2 5 3 6 7 10 5 1 2 3 4 5 2 3 4 5 6 3 4 5 6 7 4 5 6 7 8 5 6 7 8 9
 

 

Sample Output
28 46 80
 
思路：
题目大意是找两条从(1, 1) 到(n, n)的路径，使权值和最大且节点不重叠。

让两个进程同时进行，枚举步数K，当x1==x2||y1==y2时跳过，得状态转移方程：

dp(k, x1, y1, x2, y2) = max(dp(k-1, x1-1, y1, x2-1, y2), dp(k-1, x1-1, y1, x2, y2-1), dp(k-1, x1, y1-1, x2-1, y2), dp(k-1, x1, y1-1,x2, y2-1)) 

+ data(x1, y1) + data(x2, y2) ; 

由于只能走右或下，所以坐标满足x+y=k。这样就能降低维数为3维，方程：

dp(k, x1, x2) = max(dp(k-1, x1, x2), dp(k-1, x1-1, x2), dp(k-1, x1, x2-1), dp(k-1, x1-1, x2-1)) + data(x1, k-x1) + data(x2, k-x2) ;

 
代码：

#include<stdio.h>
#include<string.h>
#define MAX(a,b)   a>b?a:b
int data[31][31], dp[62][31][31] ;

int maxx(int a,int b,int c,int d)
{
    a = MAX(a,b) ;
    c = MAX(c,d) ;
    a = MAX(a,c) ;
    return a ;
}

int main()
{
    int n, i, j, k ;
    while(scanf("%d",&n)!=-1)
 {
        for(i=0; i<n; i++)
            for(j=0; j<n; j++)
                scanf("%d", &data[i][j]) ;
        memset(dp, 0, sizeof(dp)) ;
        for(k=1; k<2*n-2; k++)
        {
            for(i=0; i<=k; i++)
           {
                for(j=0; j<=k; j++)
              {
                    if(i==j||i>=n||j>=n)   
                        continue ;
                    dp[k][i][j] = maxx(dp[k-1][i][j], dp[k-1][i-1][j], dp[k-1][i][j-1], dp                             [k-1][i-1][j-1]) ;
                    dp[k][i][j] += data[i][k-i] + data[j][k-j] ;
                }
            }
        }
        dp[k][n-1][n-1] = MAX(dp[k-1][n-2][n-1], dp[k-1][n-1][n-2]) + data[n-1][n-1] + data[0][0] ;
        printf("%d\n", dp[k][n-1][n-1]) ;
    }
    return 0 ;
}

链接：http://acm.hdu.edu.cn/showproblem.php?pid=2686