【题目链接】
【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 = 1; k <= n; k++)
{
for (int i = 1; i < k; i++)
{
for (int j = i + 1; 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 = 1; i <= n; i++)
{
for (int j = 1; 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 = 0; i < res.size() - 1; i++) printf("%d ",res[i]);
printf("%d\n",res[res.size()-1]);
}
int main()
{
scanf("%d%d",&n,&m);
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
g[i][j] = mp[i][j] = INF;
pre[i][j] = i;
}
}
for (int i = 1; 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 0;
}