最小环

[poj 1734] (http://poj.org/problem?id=1734)

题目描述：
Sightseeing trip
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5500 Accepted: 2112 Special Judge
Description

There is a travel agency（中介） in Adelton town on Zanzibar island. It has decided to offer its clients（顾客）, besides many other attractions, sightseeing the town. To earn as much as possible from this attraction, the agency has accepted a shrewd（精明的，狡猾的） 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（双向公路） numbered from 1 to M. Two crossing points（交点） 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（指定） 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

题目分析：
由于是双向公路，即可以将其看成无向图，即本题要解决的问题就是找到所有路径中存在的路径最短的最小环，并把环上的每个节点都输出即可。
算法：Floyd算法拓展找最小环+回溯记录路径

代码实现：

#include <iostream>
#include <stdio.h>
using namespace std;

const int MAX=0x7fffffff;
int n,m;
int map[110][110];//记录i和j之间的距离长度
int dist[110][110];//记录i和j之间的最短路径长度
int pre[110][110];//记录他的上个节点
int path[110];//记录经过的点

int min(int x,int y)
{
    return x<=y? x:y;
}

int main()
{
    int a,b,c;
    int M;
    int num;
    M=MAX/10;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                map[i][j]=dist[i][j]=M;
                pre[i][j]=i;
            }
        }
        while(m--)
        {
            scanf("%d%d%d",&a,&b,&c);
            if(c<map[a][b])
            {
                map[a][b]=map[b][a]=c;
                dist[a][b]=dist[b][a]=c;
            }
        }
        int ans=M;
        for(int k=1; k<=n; k++)
        {
            for(int i=1; i<k; i++)
            {
                for(int j=i+1; j<k; j++)
                {
                    int pp=dist[i][j]+map[k][i]+map[k][j];
                    if(pp<ans)
                    {
                        ans=pp;
                        num=0;
                        int p=j;
                        while(p!=i)//更新他的上个节点
                        {
                            path[num++]=p;
                            p=pre[i][p];
                        }
                        path[num++]=i;
                        path[num++]=k;
                    }
                }
            }
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=n; j++)
                {
                    if(dist[i][j]>dist[k][j]+dist[i][k])//更新最短路径和当前节点的上一个节点
                    {
                        dist[i][j]=dist[i][k]+dist[k][j];
                        pre[i][j]=pre[k][j];
                    }
                }
            }
        }
        if(ans==M) printf("No solution.\n");
        else
        {
            for(int i=0; i<num; i++)
            {
                printf(i==num-1?"%d\n":"%d ",path[i]);
            }
        }
    }
    return 0;
}