poj 1734 Sightseeing trip(floyd 最小环)

poj 1734 Sightseeing trip(floyd 最小环)

题目：​​trip​​

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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

本题有这样一句话：“The road y_i (1<=i<=k-1) connects crossing points x_i and x_{i+1} ”。也就是无向图，map[a][b]=map[b][a]。寻找最大短的观光路径就是查找最小的环。因为点的数量不大，可以用O(n^3)的floyd求解。floyd: 对象可以是无向图也可以是有向图。不能有负环，基于动态规划实现。其中：

for(int k=1;k<=n;k++){

for(int i=1;i<=n;i++){

for(int j=1;j<=n;j++){

if(dis[i][k]!=INF&&dis[k][j]!=INF&&dis[i][j]>dis[i][k]+dis[k][j]){

dis[i][j]=dis[i][k]+dis[k][j];

pre[i][j]=pre[k][j];

}

}

}

}

保证了i-->j返回时不能直接j-->i。也就满足了题目要求。输出路径问题：可以在求得小环时就把涉及到的点全部放进一个队列里，每找到一个更小的环就更新自己的队列，最后就是answer。

#include #include #include using namespace std;const int maxn=105,INF=0x3f3f3f3f;int pre[maxn][maxn],map[maxn][maxn],dis[maxn][maxn];int res,n,m;void init(){ res=INF; memset(map,0x3f,sizeof(map));}int ans[maxn],top,p;void floyd(){ for(int i=1;i<=n;i++){ for(int j=1;j<=n;j++){ dis[i][j]=map[i][j]; pre[i][j]=i; } } for(int k=1;k<=n;k++){ for(int i=1;i>n>>m){ init(); for(int i=0;ic)map[a][b]=map[b][a]=c; } floyd(); if(res==INF) puts("No solution."); else { for(int i=0;i

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