poj1734Sightseeing trip

Sightseeing trip

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6811		Accepted: 2602		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

题意：给出你一个无向图，要你求一个最小环，最小环的最少有3个节点，如果有环则顺序输出节点上的点，如果没有环，输出NO solution.

分析：其实用dfs就可以AC了，但是我发现其他dalao都用floyd做，学习了一下。

     经过分析可以得到，包含i和j的最小环其实就是i到j的最短路和i到j的次短路组成的，至于为什么，很好想，因为组成了一个环，那么从i到j就有两条不同的路可以走，而要求最小环，所以就是i到j的最短路和i到j的次短路组成的。

     其实我们就是要求一点不在i到j的最短路上的k,那么最小环=min{d[i][j] + a[i][k] + a[k][j]},根据floyd算法，我们先枚举k,然后枚举i和j，不超过k即可（因为k不在i到j的最短路上），然后记录一下路径即可.路径记录可以递归，也可以就用数组记录前驱。

#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>

#define inf 0x7ffffff  

using namespace std;

int n, m,a[110][110],d[110][110],head[110][110],res,tot,ans[10010];

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
            {
            a[i][j] = d[i][j] = inf;
            head[i][j] = i;
            }
        for (int i = 1; i <= m; i++)
        {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            a[x][y] = a[y][x] = d[x][y] = d[y][x] = min(z, a[x][y]);
        }
        res = inf;
        for (int k = 1; k <= n; k++)
        {
            for (int i = 1; i < k; i++)
            {
                for (int j = i + 1; j < k; j++)
                {
                    int temp = d[i][j] + a[i][k] + a[k][j];
                    if (temp < res)
                    {
                        res = temp;
                        tot = 0;
                        int p = j;
                        while (p != i)
                        {
                            ans[++tot] = p;
                            p = head[i][p];
                        }
                        ans[++tot] = i;
                        ans[++tot] = k;
                    }
                }
            }
            for (int i = 1; i <= n; i++)
                for (int j = 1; j <= n; j++)
                    if (d[i][j] > d[i][k] + d[k][j])
                    {
                d[i][j] = d[i][k] + d[k][j];
                head[i][j] = head[k][j];
                    }
        }
        if (res == inf)
            puts("No solution.
");
        else
        {
            printf("%d", ans[1]);
            for (int i = 2; i <= tot; i++)
                printf(" %d", ans[i]);
            printf("
");
        }
    }

    return 0;
}