【POJ 1734】 Sightseeing Trip

【题目链接】

点击打开链接

【算法】

floyd求最小环

输出路径的方法如下，对于i到j的最短路，我们记pre[i][j]表示j的上一步

在进行松弛操作的时候更新pre即可

【代码】

#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXN 110
const int INF = 1e8;

int n,m;
int g[MAXN][MAXN],mp[MAXN][MAXN],pre[MAXN][MAXN];

inline void solve()
{
        int ans = INF;
        vector< int > res;
        for (int k = ; k <= n; k++)
        {
                for (int i = ; i < k; i++)
                {
                        for (int j = i + ; j < k; j++)
                        {
                                if (g[i][j] + mp[j][k] + mp[k][i] < ans)
                                {
                                        ans = g[i][j] + mp[j][k] + mp[k][i];
                                        res.clear();
                                        int tmp = j;
                                        while (tmp != i)
                                        {
                                                res.push_back(tmp);
                                                tmp = pre[i][tmp];
                                        }
                                        res.push_back(i);
                                        res.push_back(k);
                                }
                        }
                }
                for (int i = ; i <= n; i++)
                {
                        for (int j = ; j <= n; j++)
                        {
                                if (g[i][k] + g[k][j] < g[i][j])
                                {
                                        g[i][j] = g[i][k] + g[k][j];
                                        pre[i][j] = pre[k][j];
                                }
                        }
                }
        }
        if (ans == INF)
        {
                puts("No solution.");
                return;
        }
        for (int i = ; i < res.size() - ; i++) printf("%d ",res[i]);
        printf("%d\n",res[res.size()-]);
}

int main()
{

        scanf("%d%d",&n,&m);
        for (int i = ; i <= n; i++)
        {
                for (int j = ; j <= n; j++)
                {
                        g[i][j] = mp[i][j] = INF;
                        pre[i][j] = i;
                }
        }
        for (int i = ; i <= m; i++)
        {
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                if (w < g[u][v])
                        g[u][v] = mp[u][v] = g[v][u] = mp[v][u] = w;
        }
        solve();

        return ;

}