# HDU 3001 Travelling （状态压缩/三进制）

【摘要】 Travelling（HDU 3001） Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)...

## Travelling（HDU 3001）

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8931 Accepted Submission(s): 2914

Problem Description
After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn’t want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So he turns to you for help.

Input
There are several test cases,the first line is two intergers n(1<=n<=10) and m,which means he needs to visit n cities and there are m roads he can choose,then m lines follow,each line will include three intergers a,b and c(1<=a,b<=n),means there is a road between a and b and the cost is of course c.Input to the End Of File.

Output
Output the minimum fee that he should pay,or -1 if he can’t find such a route.

Sample Input
2 1
1 2 100
3 2
1 2 40
2 3 50
3 3
1 2 3
1 3 4
2 3 10

Sample Output
100
90
7

``````#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 99999999;
int n,m;
int tri = {0,1,3,9,27,81,243,729,2187,6561,19683,59049};//三进制的位权
int dig;//dig[state][k_dig]状态state的第k位是多少
int edge,dp;//dp[state][k]为状态state下最后一个点是k的最小距离
int main(){
for(int i = 0; i < 59050; i++){
int t = i;
for(int j = 1;j<=10; j++){
dig[i][j] = t%3;
t /= 3;
if(t == 0)break;
}
}
while(scanf("%d%d",&n,&m)!=EOF){
for(int i  = 1; i <= n; i++){
for(int j = 1; j <= n; j++){
edge[i][j] = MAXN;
}
}
for(int i = 1; i <= n; i++)edge[i][i] = 0;
int a,b,c;
for(int i = 1; i <= m; i++){
scanf("%d %d %d",&a,&b,&c);
if(c < edge[a][b]){
edge[a][b] = edge[b][a] = c;
}
}
for(int state = 0; state < tri[n+1]; state++){
for(int j = 1; j <= n; j++){
dp[state][j] = MAXN;
}
}
for(int i = 1; i <= n; i++){
dp[tri[i]][i] = 0;
}
int ans = MAXN;
for(int state = 0; state < tri[n+1]; state++){
int visit_all = 1;
for(int i = 1; i <= n; i++){
if(dig[state][i] == 0)visit_all = 0;
if(dp[state][i] == MAXN)continue;
for(int j = 1; j <= n; j++){
if(i == j)continue;
if(edge[i][j] == MAXN || dig[state][j] >= 2)continue;
int newState = state + tri[j];
dp[newState][j] = min(dp[newState][j],dp[state][i]+edge[i][j]);
}
}
if(visit_all){
for(int j = 1; j <= n; j++){
ans = min(ans,dp[state][j]);
}
}
}
if(ans == MAXN){
puts("-1");
continue;
}
printf("%d\n",ans);
}
}

```

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
```

【版权声明】本文为华为云社区用户转载文章，如果您发现本社区中有涉嫌抄袭的内容，欢迎发送邮件进行举报，并提供相关证据，一经查实，本社区将立刻删除涉嫌侵权内容，举报邮箱： cloudbbs@huaweicloud.com
• 点赞
• 收藏
• 关注作者

0/1000

### 设置昵称

*长度不超过10个汉字或20个英文字符，设置后3个月内不可修改。

*长度不超过10个汉字或20个英文字符，设置后3个月内不可修改。