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
1 /**/
2 #include<iostream>
3 #include<cstdio>
4 #include<cmath>
5 #include<cstring>
6 #include<algorithm>
7 using namespace std;
8 const int INF=1e6;
9 const int mxn=4194304;//2^22
10 int dp[mxn][22];//[遍历状态][最后到达点]=最短路径
11 int dis[22][22];
12 int pre[22];//每个点的前驱要求
13 int n,m;
14 int xn;
15 int main(){
16 while(scanf("%d%d",&n,&m)!=EOF){
17 memset(pre,0,sizeof pre);
18 int i,j;
19 xn=1<<n;
20 for(i=1;i<xn;i++)
21 for(j=0;j<n;j++){
22 dp[i][j]=INF;
23 }
24 //init
25 for(i=0;i<n;i++)
26 for(j=0;j<n;j++){
27 scanf("%d",&dis[i][j]);
28 if(dis[i][j]==-1)dis[i][j]=INF;
29 }
30 int u,v;
31 for(i=1;i<=m;i++){//保存前驱要求
32 scanf("%d%d",&u,&v);
33 pre[v]|=(1<<u);
34 }
35 dp[1][0]=0;
36 for(i=1;i<xn;i++){
37 for(j=0;j<n;j++){
38 if(dp[i][j]==INF)continue;//i状态之前没走到
39 for(int k=1;k<n;k++){
40 if(!(i&(1<<j)))continue;//j不在已走过的集合中
41 if(i&(1<<k))continue;//k在走过的集合中
42 if(pre[k]!=(i&pre[k]))continue;//k点前驱要求未满足
43 dp[i|(1<<k)][k]=min(dp[i|(1<<k)][k],dp[i][j]+dis[j][k]);
44 }
45 }
46 }
47 int ans=INF;
48 for(i=0;i<n;i++)ans=min(ans,dp[xn-1][i]);
49 if(ans>=INF) printf("-1\n");
50 else printf("%d\n",ans);
51 }
52 return 0;
53 }