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);

	}

}