DFS HDOJ 2181 哈密顿绕行世界问题

题目传送门

题意:中文题面

分析:直接排完序后DFS.这样的题以后不应该再写题解的.

#include <bits/stdc++.h>
using namespace std;

vector<int> G[21];
int ans[21];
int v[3];
bool vis[21];
int cnt;

void print()	{
	printf ("%d:  ", ++cnt);
	for (int i=0; i<20; ++i)	{
		printf ("%d ", ans[i]);
	}
	printf ("%d\n", ans[20]);
}

void DFS(int u, int x, int step)	{
	for (int i=0; i<G[u].size (); ++i)	{
		int v = G[u][i];
		if (v == x)	{
			if (step < 20)	continue;
			ans[step] = x;
			print ();	return ;
		}
		else if (vis[v])	continue;
		vis[v] = true;
		ans[step] = v;
		DFS (v, x, step + 1);
		vis[v] = false;
	}
}

int main(void)	{
	for (int i=1; i<=20; ++i)	{
		for (int j=0; j<3; ++j)	{
			scanf ("%d", &v[j]);
			G[i].push_back (v[j]);
			//G[v[j]].push_back (i);
		}
	}
	for (int i=1; i<=20; ++i)	{
		sort (G[i].begin (), G[i].end ());
	}
	int m;
	while (scanf ("%d", &m) == 1)	{
		if (!m)	break;
		cnt = 0;
		memset (vis, false, sizeof (vis));
		vis[m] = true;
		ans[0] = m;
		DFS (m, m, 1);
	}

	return 0;
}