Sightseeing trip
Time Limit: 1000MS Memory Limit: 65536K
Special Judge
Description

There is a travel agency in Adelton town on Zanzibar island. It has decided to offer its clients, besides many other attractions, sightseeing the town. To earn as much as possible from this attraction, the agency has accepted a shrewd decision: it is necessary to find the shortest route which begins and ends at the same place. Your task is to write a program which finds such a route.

In the town there are N crossing points numbered from 1 to N and M two-way roads numbered from 1 to M. Two crossing points can be connected by multiple roads, but no road connects a crossing point with itself. Each sightseeing route is a sequence of road numbers y_1, ..., y_k, k>2. The road y_i (1<=i<=k-1) connects crossing points x_i and x_{i+1}, the road y_k connects crossing points x_k and x_1. All the numbers x_1,...,x_k should be different.The length of the sightseeing route is the sum of the lengths of all roads on the sightseeing route, i.e. L(y_1)+L(y_2)+...+L(y_k) where L(y_i) is the length of the road y_i (1<=i<=k). Your program has to find such a sightseeing route, the length of which is minimal, or to specify that it is not possible,because there is no sightseeing route in the town.
Input

The first line of input contains two positive integers: the number of crossing points N<=100 and the number of roads M<=10000. Each of the next M lines describes one road. It contains 3 positive integers: the number of its first crossing point, the number of the second one, and the length of the road (a positive integer less than 500).
Output

There is only one line in output. It contains either a string ‘No solution.‘ in case there isn‘t any sightseeing route, or it contains the numbers of all crossing points on the shortest sightseeing route in the order how to pass them (i.e. the numbers x_1 to x_k from our definition of a sightseeing route), separated by single spaces. If there are multiple sightseeing routes of the minimal length, you can output any one of them.
Sample Input

5 7
1 4 1
1 3 300
3 1 10
1 2 16
2 3 100
2 5 15
5 3 20
Sample Output

1 3 5 2
Source

CEOI 1999

題意：給你一張無向圖，求圖中的最小環，輸出路徑。

題解：

floyed求最小環

詳見代碼。

 1 #include<cstdio>    
 2 #include<cstring>    
 3 #define find_min(a,b) a<b?a:b    
 4     
 5 const int N=109;    
 6 const int INF=0x7ffffff;    

 7 int f[N][N],dis[N][N],pre[N][N],path[N],n,m,num;  
 8     
 9 int main()    
10 {    
11         scanf("%d%d",&n,&m);
12       
13         for(int i=1;i<=n;i++)   
14             for(int j=1;j<=n;j++) 
15                 f[i][j]=dis[i][j]=INF,pre[i][j]=i;  
16                 
17         while(m--)
18         {    
19             int a,b,c;
20             scanf("%d%d%d",&a,&b,&c);    
21             f[a][b]=f[b][a]=dis[a][b]=dis[b][a]=find_min(f[a][b],c);    
22         }    
23     
24         int min=INF;    
25         for(int k=1;k<=n;k++)//最短路徑外一點將最短路首尾鏈接,那麽就得到一個最小環,枚舉中間點  
26         {  
27             for(int i=1;i<k;i++)   
28                 for(int j=i+1;j<k;j++)
29                 {    
30                     //求最小環不能用兩點間最短路松弛,因為(i,k)之間的最短路,(k,j)之間的最短路可能有重合的部分    
31                     //所以f[][]其實是不更新的,這裏和單純的floyd最短路不一樣    
32                     //dis[i][j]保存的是 i 到 j 的最短路權值和,(i->j)    
33                     int tmp=dis[i][j]+f[i][k]+f[k][j];//這裏k分別和i還有 j在f[][]中直接相連    
34                     if(tmp<min)
35                     {    
36                         min=tmp;num=0;    
37                         int p=j;    
38                         while(p!=i)//回溯  
39                         {  
40                             path[num++]=p;    
41                             p=pre[i][p];//pre[i][j]表示i->j最短路徑上j前面的一個點    
42                         }    
43                         path[num++]=i; path[num++]=k;    
44                     }    
45                 }  
46             
47             for(int i=1;i<=n;i++)   
48                 for(int j=1;j<=n;j++)   
49                     if(dis[i][j]>dis[i][k]+dis[k][j])
50                     {    
51                         dis[i][j]=dis[i][k]+dis[k][j];//dis[][]保存兩點間最短距離    
52                         pre[i][j]=pre[k][j];    
53                     }    
54                 
55         }    
56         if(min==INF)puts("No solution.");    
57         else
58         {    
59             printf("%d",path[0]);    
60             for(int i=1;i<num;i++)    
61                 printf(" %d",path[i]);    
62             puts("");    
63         }  
64     return 0;    
65 }    

Sightseeing trip