POJ 2942.Knights of the Round Table 解题报告

/*
       求有向图的点双连通分支tarjan算法
       思路：
       1.对图先进行深度优先搜索形成搜索数，计算每一个节点的先深编号dfn[n]
       2.计算所有节点v的low[v]是在先深生成树上按照后根遍历的顺序进行的.
          因此,当仿问节点v时它的每一个儿子u的low[u]已经计算完毕这时low[v]取下面三值的最小者:
          1)dfn[v];
          2)dfn[w],对于回退边(v,w)
          3)low[u],对于v的任何儿子u
       3.判断一个顶点是不是桥,割点:
         a)v为树根,且v有多于1个子树
         b)v不为树根,且满足存在边(v,u) ,使得dfn[v]<=low[u].
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 1009
using namespace std;
int n, m, x, y;
bool g[INF][INF];
int low[INF], dfn[INF], sta[INF], ans[INF][INF], f[INF], ok[INF], Top, Max , tcc, t;
bool make (int x, int cow) {
	for (int i = 0; i < cow; i++) {
		int v = ans[tcc][i];
		if (x != v && g[x][v]) {
			if (f[v] == -1) {
				f[v] = !f[x];
				if (make (v, cow) ) return 1;
			}
			else if (f[v] == f[x]) return 1;
		}
	}
	return 0;
}
void check (int cow) {
	memset (f, -1, sizeof f);
	f[ans[tcc][0]] = 0;
	if (make (ans[tcc][0], cow) )
		for (int i = 0; i < cow; i++)
			ok[ans[tcc][i]] = 1;
}
void dfs (int k, int from) {
	sta[++Top] = k;
	low[k] = dfn[k] = ++t;
	for (int i = 1; i <= n; i++) {
		if (i == from || g[k][i] == 0 || k == i) continue;
		if (!dfn[i]) {
			dfs (i, k);
			low[k] = min (low[k], low[i]);
			if (dfn[k] <= low[i]) {
				ans[tcc][0] = k;
				int cow = 1;
				do
					ans[tcc][cow++] = sta[Top];
				while (sta[Top--] != i);
				if (cow > 2) check (cow), ++tcc;
			}
		}
		else low[k] = min (low[k], dfn[i]);
	}
	return ;
}
void Tarjan (int n) {
	memset (low, 0, sizeof low);
	memset (dfn, 0, sizeof dfn);
	Top = tcc = t = 0;
	for (int i = 1; i <= n; i++)
		if (dfn[i] == 0) dfs (i, -1);
}
int main() {
	while (~scanf ("%d %d", &n, &m) ) {
		if (n == 0 && m == 0) return 0;
		memset (g, 1, sizeof g);
		memset (ok, 0, sizeof ok);
		Max = 0;
		for (int i = 1; i <= m; i++) {
			scanf ("%d %d", &x, &y);
			g[x][y] = g[y][x] = 0;
		}
		Tarjan (n);
		int ans = 0;
		for (int i = 1; i <= n; i++)
			if (!ok[i]) ans++;
		printf ("%d\n", ans);
	}
	return 0;
}