Poj 1734 Sightseeing trip floyd最小环

Sightseeing trip
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5698 Accepted: 2203 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

题意：给n个点，m条双向边，找出最小环。
找出最小环可看成i->j的路径可由i->k->j->i这条更短的路径更新而来，可以推出一个路径长度表达式dis[i][j]+Map[i][k]+Map[k][j]。因为是最小环，不必考虑方向，即不考虑谁指向谁，只要是个环就行了（之前一直在这里纠结，真是无药可救），即可以采用floyd来跑找出最小环。因为floyd是个更新最短路的过程，更新过后环路径可能就不存在了，所以必须在更新前找出最小环，所以查找过程写在更新最短路之前。定义一个ans=inf，由如果ans>dis[i][j]+Map[i][k]+Map[k][j]，则更新ans，如果不能更新就直接输出”No solution.”。能更新则用findpath()来储存路径。还有在更新最短路时，最短路径由i->j变为i->k->j，所以path[i][j]要更新为path[k][j]。

#include"stdio.h"#include"iostream"#include"algorithm"#include"string.h"#define maxn 510#define inf 0xfffffffusing namespace std;int n,m;int num;int dis[maxn][maxn];  ///i->j的最短路int Map[maxn][maxn];  ///图int path[maxn][maxn];  ///用更新最短时储存路径和更新road用的int road[maxn];  ///路径void init()   ///初始化{    for(int i = 0;i < maxn;i++)    {        road[i] = inf;        for(int j = 0;j < maxn;j++)        {            dis[i][j] = Map[i][j] = inf;            path[i][j] = i;        }    }}void findpath(int i,int j,int k)    ///储存路径，大概为j->i->k->(j,输出环不用输出起点){    for(num = 0;i-j;j = path[i][j])        road[num++] = j;    road[num++] = i;    road[num++] = k;}int Floyed(){    int ans = inf;    for(int k = 1;k <= n;k++)    {        for(int i = 1;i < k;i++)      ///k点不属于i->j的最短路一条，所以i<k            for(int j = i+1;j < k;j++)            {                if(ans > dis[i][j]+Map[i][k]+Map[k][j])   ///更新最小环的值                {                    ans = dis[i][j]+Map[i][k]+Map[k][j];                    findpath(i,j,k);                }            }        for(int i = 1;i <= n;i++)   ///更新最短路            for(int j = 1;j <= n;j++)                if(dis[i][j] > dis[i][k]+dis[k][j])                {                    dis[i][j] = dis[i][k]+dis[k][j];                    path[i][j] = path[k][j];                }    }    return ans;}int main(void){    while(scanf("%d%d",&n,&m) !=EOF)    {        init();        for(int i = 1;i <= m;i++)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            if(Map[a][b] > c)            {                Map[a][b] = dis[a][b] = Map[b][a] = dis[b][a] = c;   ///双向边            }        }        int ans = Floyed();        if(ans == inf)            printf("No solution.\n");        else        {            for(int i = 0;i < num;i++)   ///输出路径            {                if(i < num-1)                    printf("%d ",road[i]);                else                    printf("%d",road[i]);            }            printf("\n");        }    }    return 0;}