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POJ 2438 解题报告

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);	}}