HDU 2686 Matrix (多线程DP)

Matrix

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1005    Accepted Submission(s): 558

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

 

 

Author

yifenfei

 

 

Source

ZJFC 2009-3 Programming Contest

 

 

Recommend

yifenfei

 

 

多进程DP，昨天第一次听说...

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) ;

 

 

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 
 6 using namespace std;
 7 
 8 int data[40][40],dp[80][40][40];
 9 
10 int Max(int a,int b,int c,int d){
11     return max(a,max(b,max(c,d)));
12 }
13 
14 int main(){
15 
16     //freopen("input.txt","r",stdin);
17 
18     int n;
19     while(~scanf("%d",&n)){
20         int i,j,k;
21         for(i=0;i<n;i++)
22             for(j=0;j<n;j++)
23                 scanf("%d",&data[i][j]);
24         memset(dp,0,sizeof(dp));
25         for(k=1;k<2*n-2;k++)
26             for(i=0;i<n;i++)
27                 for(j=0;j<n;j++){
28                     if(i==j)
29                         continue;
30                     dp[k][i][j]=Max(dp[k-1][i][j],dp[k-1][i-1][j],dp[k-1][i][j-1],dp[k-1][i-1][j-1]);
31                     dp[k][i][j]+=data[i][k-i]+data[j][k-j];
32                 }
33         int ans=max(dp[k-1][n-1][n-2],dp[k-1][n-2][n-1])+data[0][0]+data[n-1][n-1];
34         printf("%d\n",ans);
35     }
36     return 0;
37 }