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 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
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 }