poj 2942 Knights of the Round Table 双连通分量

白书上的代码

#include<cstdio>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>

using namespace std;

struct Edge { int u, v; };

const int maxn = 1000 + 10;
int pre[maxn], iscut[maxn], bccno[maxn], dfs_clock, bcc_cnt; // 割顶的bccno无意义
vector<int> G[maxn], bcc[maxn];

stack<Edge> S;

int dfs(int u, int fa) 
{
	int lowu = pre[u] = ++dfs_clock;
	int child = 0;
	for(int i = 0; i < G[u].size(); i++) 
	{
		int v = G[u][i];
		Edge e;
		e.u=u;
		e.v=v;
		if(!pre[v]) 
		{ // 没有访问过v
			S.push(e);
			child++;
			int lowv = dfs(v, u);
			lowu = min(lowu, lowv); // 用后代的low函数更新自己
			if(lowv >= pre[u]) 
			{
				iscut[u] = true;
				bcc_cnt++; bcc[bcc_cnt].clear();
				for(;;) 
				{
					Edge x = S.top(); S.pop();
					if(bccno[x.u] != bcc_cnt) { bcc[bcc_cnt].push_back(x.u); bccno[x.u] = bcc_cnt; }
					if(bccno[x.v] != bcc_cnt) { bcc[bcc_cnt].push_back(x.v); bccno[x.v] = bcc_cnt; }
					if(x.u == u && x.v == v) break;
				}
			}
		}
		else if(pre[v] < pre[u] && v != fa) {
		S.push(e);
		lowu = min(lowu, pre[v]); // 用反向边更新自己
		}
	}
	if(fa < 0 && child == 1) iscut[u] = 0;
	return lowu;
}

void find_bcc(int n) 
{
	// 调用结束后S保证为空，所以不用清空
	memset(pre, 0, sizeof(pre));
	memset(iscut, 0, sizeof(iscut));
	memset(bccno, 0, sizeof(bccno));
	dfs_clock = bcc_cnt = 0;
	for(int i = 0; i < n; i++)
    if(!pre[i]) dfs(i, -1); 
}

int odd[maxn], color[maxn];
bool bipartite(int u, int b) 
{
	for(int i = 0; i < G[u].size(); i++) 
	{
		int v = G[u][i]; if(bccno[v] != b) continue;
		if(color[v] == color[u]) return false;
		if(!color[v]) 
		{
			color[v] = 3 - color[u];
			if(!bipartite(v, b)) return false;
		}
	}
	return true;
}

int A[maxn][maxn];

int main() 
{
	int kase = 0, n, m;
	while(scanf("%d%d", &n, &m) == 2 && n) 
	{
		for(int i = 0; i < n; i++) G[i].clear();
		memset(A, 0, sizeof(A));
		for(int i = 0; i < m; i++) 
		{
			int u, v;
			scanf("%d%d", &u, &v); u--; v--; A[u][v] = A[v][u] = 1;
		}
		for(int u = 0; u < n; u++)
			for(int v = u+1; v < n; v++)
				if(!A[u][v]) { G[u].push_back(v); G[v].push_back(u); }

		find_bcc(n);

		memset(odd, 0, sizeof(odd));
		for(int i = 1; i <= bcc_cnt; i++) 
		{
			memset(color, 0, sizeof(color));
			for(int j = 0; j < bcc[i].size(); j++) bccno[bcc[i][j]] = i; // 主要是处理割顶
			int u = bcc[i][0];
			color[u] = 1;
			if(!bipartite(u, i))
			for(int j = 0; j < bcc[i].size(); j++) odd[bcc[i][j]] = 1;
		}
		int ans = n;
		for(int i = 0; i < n; i++) if(odd[i]) ans--;
		printf("%d
", ans);
	}
	return 0;
}