状压DP HDU3538 A sample Hamilton path

A sample Hamilton path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 527    Accepted Submission(s): 213

Problem Description

Give you a Graph,you have to start at the city with ID zero.

 

 

Input

The first line is n(1<=n<=21) m(0<=m<=3)
The next n line show you the graph, each line has n integers.
The jth integers means the length to city j.if the number is -1 means there is no way. If i==j the number must be -1.You can assume that the length will not larger than 10000
Next m lines,each line has two integers a,b (0<=a,b<n) means the path must visit city a first.
The input end with EOF.

 

 

Output

For each test case,output the shorest length of the hamilton path.
If you could not find a path, output -1

 

 

Sample Input

3 0-1 2 4-1 -1 21 3 -14 3-1 2 -1 12 -1 2 14 3 -1 13 2 3 -11 30 12 3

 

 

Sample Output

45

Hint

I think that all of you know that a!=b and b!=0 =。=

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（11）——Host by BUPT

 

 

Recommend

zhouzeyong

 

 

莫名其妙就状压DP了，说实在我现在还不明白这是个什么玩意儿......

不过尹神说了是典型，那就写一写吧......

这就是个Floyd么......

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6  7 int n,m; 8 int dis[25][25]; 9 int f[2500000];10 int dp[2500000][25];//dp[i][j]已访问的点集i 到达点i的最短hamilton 11 const int MAX=3000000;12 int ans;13 14 int main(){15     int x=0,y=0;16     while(scanf("%d%d",&n,&m)!=EOF){17         memset(f,0,sizeof(f));18         memset(dis,0,sizeof(dis));19         memset(dp,0,sizeof(dp)); 20         for(int i=0;i<n;i++)21             for(int j=0;j<n;j++) scanf("%d",&dis[i][j]);22         for(int i=1;i<=m;i++){23             scanf("%d%d",&x,&y);24             f[y]|=(1<<x);//观察了样例才发现 原来y可以多次出现 25         }26         for(int i=0;i<(1<<n);i++)27             for(int j=0;j<n;j++)  dp[i][j]=MAX;28         dp[1][0]=0;29         for(int k=0;k<(1<<n);k++)30             for(int i=0;i<n;i++)31                 if(dp[k][i]!=MAX)32                     for(int j=0;j<n;j++){33                         if((dis[i][j]==-1)||(!(k&(1<<i)))||(k&(1<<j))||f[j]!=(k&f[j])) continue;//判断是否遍历过 34                         dp[k|(1<<j)][j]=min(dp[k|(1<<j)][j],dp[k][i]+dis[i][j]);//f[k][i]->f[k+{j}][j] + dis(i, j) 35                     }36         ans=MAX;37         for(int i=0;i<n;i++) ans=min(ans,dp[(1<<n)-1][i]);38         if(ans==MAX) printf("-1\n");39         else printf("%d\n",ans);40     }41     return 0;42 }