POJ 1734：Sightseeing trip

Sightseeing trip
Time Limit: 1000MS        Memory Limit: 65536K
Total Submissions: 6831        Accepted: 2612        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

题目

　　（PS：其实就是求图上最小环啦）

　　芒果君：本来以为自己最短路学的可以来着，结果知道最小环用floyd而不用tarjan时我的内心是崩溃的，然后也打不出来。这道题的巧妙之处在于，求环的过程和floyd一块做而在其之前，使得不会在结果中出现重复节点。最短路无非是加了一句记录中转点；求环的话每次都要重做，首先要清楚它不只是一条最短路，还有一个不在路上的点k将其首尾相连，先记录i，再进行递归找到最短路上更新的所有点，这段代码需要仔细理解。

　 感觉这道题不太好想呢？

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #define inf 1<<29 
 7 using namespace std;
 8 int ans,dis[110][110],road[110][110],ma[110][110],path[110],n,m,cnt; 
 9 int read()
10 {
11     int x=0;
12     char ch=getchar();
13     while(ch<'0'||ch>'9') ch=getchar();
14     while(ch>='0'&&ch<='9'){
15         x=x*10+ch-'0';
16         ch=getchar(); 
17     }
18     return x; 
19 }
20 void init()
21 {
22     ans=inf; 
23     memset(road,0,sizeof(road));
24     memset(ma,0,sizeof(ma)); 
25     for(int i=1;i<=n;++i)
26         for(int j=1;j<=n;++j)
27             dis[i][j]=inf;
28 }
29 void record(int x,int y)
30 {
31     if(road[x][y]){
32         record(x,road[x][y]);
33         record(road[x][y],y); 
34     }
35     else path[++cnt]=y;
36 }
37 int main(){
38     int x,y,t,i,j,k; 
39     while((scanf("%d%d",&n,&m))!=EOF){
40         init(); 
41         for(i=1;i<=m;++i){
42             x=read();y=read();t=read();
43             if(t<dis[x][y]) dis[x][y]=dis[y][x]=t;  
44         }
45         for(i=1;i<=n;++i)
46             for(j=1;j<=n;++j)
47                 ma[i][j]=dis[i][j]; 
48         for(k=1;k<=n;++k){
49             for(i=1;i<k;++i)
50                 for(j=i+1;j<k;++j){
51                     if(ans>dis[i][j]+ma[i][k]+ma[k][j]){
52                         ans=dis[i][j]+ma[i][k]+ma[k][j];
53                         cnt=0;
54                         path[++cnt]=i;
55                         record(i,j); 
56                         path[++cnt]=k; 
57                     }
58                 }
59             for(i=1;i<=n;++i)
60                 for(j=1;j<=n;++j)
61                     if(dis[i][j]>dis[i][k]+dis[k][j]){
62                         dis[i][j]=dis[i][k]+dis[k][j];
63                         road[i][j]=k; 
64                     }
65         }
66         if(ans==inf) printf("No solution.\n");
67         else{
68             for(i=1;i<=cnt;i++) printf("%d ",path[i]); 
69             printf("\n");
70         } 
71     }
72     return 0;
73 }

 

 

转载于:https://www.cnblogs.com/12mango/p/7203441.html