【双连通分量】 POJ Knights of the Round Table

白书上的例题。。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1005
#define maxm 2000005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

struct Edge
{
	int v;
	Edge *next;
}E[maxm], *H[maxn], *edges;

stack<int> s;
queue<int> q;

int g[maxn][maxn];
int color[maxn];
int mark[maxn];
int dfn[maxn];
int low[maxn];
int odd[maxn];
int dfs_clock;
int n, m;

void addedges(int u, int v)
{
	edges->v = v;
	edges->next = H[u];
	H[u] = edges++;
}

void init(void)
{
	dfs_clock = 0;
	edges = E;
	memset(H, 0, sizeof H);
	memset(dfn, 0, sizeof dfn);
	memset(low, 0, sizeof low);
	memset(odd, 0, sizeof odd);
	for(int i = 0; i <= n; i++)
		for(int j = 0; j <= n; j++)
			g[i][j] = 0;
	while(!s.empty()) s.pop();
}

void read(void)
{
	int u, v;
	while(m--) {
		scanf("%d%d", &u, &v);
		g[u][v] = g[v][u] = 1;
	}
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			if(!g[i][j] && i != j)
				addedges(i, j);
}

bool check(int s)
{
	memset(color, 0, sizeof color);
	int cnt = 0;
	for(int i = 1; i <= n; i++) cnt += mark[i];
	if(cnt < 3) return true;
	while(!q.empty()) q.pop();
	q.push(s);
	color[s] = 1;
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		for(Edge *e = H[u]; e; e = e->next) {
			int v = e->v;
			if(!mark[v]) continue;
			if(color[u] == color[v]) return false;
			if(!color[v]) {
				color[v] = 3 - color[u];
				q.push(v);
			}
		}
	}
	return true;
}

void solve(int u)
{
	if(!check(u))
		for(int i = 1; i <= n; i++)
			if(mark[i]) odd[i] = 1;
}

void tarjan(int u, int fa)
{
	dfn[u] = low[u] = ++dfs_clock;
	s.push(u);
	for(Edge *e = H[u]; e; e = e->next) {
		int v = e->v;
		if(!dfn[v]) {
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
			if(low[v] >= dfn[u]) {
				int vn;
				memset(mark, 0, sizeof mark);
				do
				{
					vn = s.top();
					s.pop();
					mark[vn] = 1;
				}while(vn != v);
				mark[u] = 1;
				solve(u);
			}

		}
		else low[u] = min(low[u], dfn[v]);
	}
}

void work(void)
{
	int ans = 0;
	for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i, i);
	for(int i = 1; i <= n; i++) if(!odd[i]) ans++;
	printf("%d\n", ans);
}

int main(void)
{
	while(scanf("%d%d", &n, &m), n != 0 || m != 0) {
		init();
		read();
		work();
	}

	return 0;
}