POJ 2438 （哈密顿回路）

分析:

       2*n个小朋友,每个最多有n-1个"敌人",显然是存在哈密顿回路的.

       预处理边,然后找哈密顿回路.

code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
#define pb push_back
#define sz(a)  (int)(a).size()

const int INF = 500;
bool edge[INF][INF];
typedef vector<int> vi;
vi ans;
//求哈密顿回路O(n^2)
void Hamilton (vi& ans, bool edge[INF][INF], int n) {
	int s = 1, tol = 2, t, i, j;
	bool vis[INF] = {0};
	for (i = 1; i <= n; i++) if (edge[s][i]) break;
	t = i;
	vis[s] = vis[t] = 1;
	ans.pb (s); ans.pb (t);
	while (1) {
		//头尾拓展
		while (1) {
			for (i = 1; i <= n; i++) {
				if (edge[t][i] && !vis[i]) {
					vis[i] = 1; t = i;
					ans.pb (i);
					break;
				}
			}
			if (i > n) break;
		}
		reverse (ans.begin(), ans.end() );
		swap (s, t);
		while (1) {
			for (i = 1; i <= n; i++) {
				if (edge[t][i] && !vis[i]) {
					vis[i] = 1; t = i;
					ans.pb (i);
					break;
				}
			}
			if (i > n) break;
		}
		//如果S和T不相连
		if (!edge[s][t]) {
			for (i = 1; i < sz (ans) - 2; i++)
				if (edge[ans[i]][t] && edge[ans[i + 1]][s]) break;
			reverse (ans.begin() + i + 1, ans.end() );
			t = * (ans.end() - 1);
		}
		tol = sz (ans);
		if (tol == n) return;
		//如果还有点未加入ans
		for (j = 1; j <= n; j++) {
			if (vis[j]) continue;
			//找到与这个点相连的点
			for (i = 1; i < tol - 1; i++) if (edge[ans[i]][j]) break;
			if (edge[ans[i]][j]) break;
		}
		s = ans[i - 1], t = j;
		reverse (ans.begin(), ans.begin() + i );
		reverse (ans.begin() + i, ans.end() );
		ans.pb (j), vis[j] = 1;
	}
}
int n, m;
int main() {
	while (~scanf ("%d %d", &n, &m) ) {
		if (n == 0 && m == 0) return 0;
		memset (edge, 1, sizeof edge);
		for (int i = 1; i <= 2 * n; i++) edge[i][i] = 0;
		int x, y;
		for (int i = 1; i <= m; i++) {
			scanf ("%d %d", &x, &y);
			edge[y][x] = edge[x][y] = 0;
		}
		ans.clear();
		Hamilton (ans, edge, n << 1);
		printf ("%d", ans[0]);
		for (int i = 1; i < sz (ans); i++)
			printf (" %d", ans[i]);
		putchar (10);
	}
}