HDU3538 A sample Hamilton path

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 503    Accepted Submission(s): 200

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 2
1 3 -1
4 3
-1 2 -1 1
2 -1 2 1
4 3 -1 1
3 2 3 -1
1 3
0 1
2 3

 

Sample Output

4
5

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

 

 

状态压缩DP，详解见代码

 /**/
 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int INF=1e6;
 const int mxn=;//2^22
 int dp[mxn][];//[遍历状态][最后到达点]=最短路径
 int dis[][];
 int pre[];//每个点的前驱要求
 int n,m;
 int xn;
 int main(){
     while(scanf("%d%d",&n,&m)!=EOF){
         memset(pre,,sizeof pre);
         int i,j;
         xn=<<n;
         for(i=;i<xn;i++)
          for(j=;j<n;j++){
              dp[i][j]=INF;
          }
         //init
         for(i=;i<n;i++)
          for(j=;j<n;j++){
              scanf("%d",&dis[i][j]);
              if(dis[i][j]==-)dis[i][j]=INF;
          }
         int u,v;
         for(i=;i<=m;i++){//保存前驱要求
             scanf("%d%d",&u,&v);
             pre[v]|=(<<u);
         }
         dp[][]=;
         for(i=;i<xn;i++){
             for(j=;j<n;j++){
                 if(dp[i][j]==INF)continue;//i状态之前没走到
                 for(int k=;k<n;k++){
                     if(!(i&(<<j)))continue;//j不在已走过的集合中
                     if(i&(<<k))continue;//k在走过的集合中
                     if(pre[k]!=(i&pre[k]))continue;//k点前驱要求未满足
                     dp[i|(<<k)][k]=min(dp[i|(<<k)][k],dp[i][j]+dis[j][k]);
                 }
             }
         }
         int ans=INF;
         for(i=;i<n;i++)ans=min(ans,dp[xn-][i]);
         if(ans>=INF) printf("-1\n");
         else printf("%d\n",ans);
     }
     return ;
 }