Sightseeing trip

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions:8588		Accepted:3224		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：

有n個點，m條帶權無向邊。希望找到一條路徑最短的環，環上至少包含三個點，輸出長度和路徑。

思路：

第一次寫這種最小環問題。其實本質就是要列舉節點，再列舉路徑上和他直接相鄰的兩個節點。

在floyd演算法的步驟裡，外層迴圈k剛開始時，$d[i,j]$儲存的是從$i$經過$1~k-1$中的某個節點到$j$的最短路長度。

那麼$d[i,j] + g[j,k] + g[k,i]$就是一個從$i$經過$j$直接到$k$再直接回到$i$的一個環的路徑，這個的最小值就是他對應的路徑的最小環。

列舉每一個$k$，找到上面式子的最小值，就是整個圖的最小環。

輸出路徑的話其實就是相當於一個dfs

注意點：

有重邊，所以存的是邊的最小值。

$d[i,j] + g[j,k] + g[k,i]$可能會在運算時超出int，因為可能有的兩兩節點之間沒有邊存在也就是inf

 

虐狗寶典閱讀筆記：

對於有向圖的最小環問題，可列舉起點$s = 1~n$，執行堆優化的Dijkstra演算法求解單源最短路經。$s$一定是第一個被從堆中取出的節點，我們掃描$s$的所有出邊，當擴充套件、更新完成後，令$d[s] = inf$,然後繼續求解。當$s$第二次被從堆中取出時，$d[s]就是經過點$s$的最小環長度。

 1 #include<iostream>
 2 //#include<bits/stdc++.h>
 3 #include<cstdio>
 4 #include<cmath>
 5 #include<cstdlib>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<vector>
10 #include<set>
11 #include<climits>
12 using namespace std;
13 typedef long long LL;
14 #define N 100010
15 #define pi 3.1415926535
16 #define inf 0x3f3f3f3f
17 
18 int n, m;
19 const int maxn = 105;
20 int g[maxn][maxn], d[maxn][maxn], pos[maxn][maxn], ans;
21 vector<int>path;
22 
23 void get_path(int i, int j)
24 {
25     if(pos[i][j] == 0)return;
26     get_path(i, pos[i][j]);
27     path.push_back(pos[i][j]);
28     get_path(pos[i][j], j);
29 }
30 
31 void floyd()
32 {
33     ans = inf;
34     memcpy(d, g, sizeof(g));
35     for(int k = 1; k <= n; k++){
36         for(int i = 1; i < k; i++){
37             for(int j = i + 1; j < k; j++){
38                 if(ans > (long long)d[i][j] + g[j][k] + g[k][i]){
39                     ans = d[i][j] + g[j][k] + g[k][i];
40                     path.clear();
41                     path.push_back(i);
42                     get_path(i, j);
43                     path.push_back(j);
44                     path.push_back(k);
45                 }
46             }
47         }
48         for(int i = 1; i <= n; i++){
49             for(int j = 1; j <= n; j++){
50                 if(d[i][j] > d[i][k] + d[k][j]){
51                     d[i][j] = d[i][k] + d[k][j];
52                     pos[i][j] = k;
53                 }
54             }
55         }
56     }
57 }
58 
59 int main()
60 {
61     while(scanf("%d%d", &n, &m) != EOF){
62         memset(g, 0x3f, sizeof(g));
63         memset(pos, 0, sizeof(pos));
64         for(int i = 1; i <= n; i++){
65             g[i][i] = 0;
66         }
67         for(int i = 0; i < m; i++){
68             int u, v, w;
69             scanf("%d%d%d", &u, &v, &w);
70             g[u][v] = g[v][u] = min(g[u][v], w);
71         }
72 
73         floyd();
74         if(ans == inf){
75             printf("No solution.\n");
76         }
77         else{
78             for(int i = 0; i < path.size(); i++){
79                 printf("%d", path[i]);
80                 if(i == path.size() - 1){
81                     printf("\n");
82                 }
83                 else{
84                     printf(" ");
85                 }
86             }
87         }
88     }
89     return 0;
90 }