【HDU3394】Railway【BCC】【桥】

题意：

n个点的无向图，求桥边个数，求属于多个环的边的个数。

BCC裸题？

跑一次tarjan都可以求出来啦。

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

using namespace std;

const int maxn = 10005, maxm = 100005;

int n, m, head[maxn], cnt, low[maxn], dfn[maxn], ans1, ans2, clo;
bool inbcc[maxn];

struct _edge {
	int v, next;
} g[maxm << 1];

stack<int> sta;
vector<int> bcc;

inline int iread() {
	int f = 1, x = 0; char ch = getchar();
	for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;
	for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
	return f * x;
}

inline void add(int u, int v) {
	g[cnt] = (_edge) {v, head[u]};
	head[u] = cnt++;
}

inline void update() {
	for(int i = 1; i <= n; i++) inbcc[i] = 0;
	for(int i = 0; i < bcc.size(); i++) inbcc[bcc[i]] = 1;
	int tot = 0;
	for(int j = 0; j < bcc.size(); j++) for(int i = head[bcc[j]]; ~i; i = g[i].next) if(inbcc[g[i].v])
		tot++;
	tot >>= 1;
	if(tot > bcc.size()) ans2 += tot;
}

void tarjan(int x, int f) {
	low[x] = dfn[x] = ++clo;
	sta.push(x);
	for(int i = head[x]; ~i; i = g[i].next) {
		int v = g[i].v;
		if(v == f) continue;
		if(!dfn[v]) {
			tarjan(v, x);
			low[x] = min(low[x], low[v]);
			if(low[v] > dfn[x]) ans1++;
			if(low[v] >= dfn[x]) {
				bcc.clear();
				while(1) {
					int u = sta.top(); sta.pop();
					bcc.push_back(u);
					if(u == v) break;
				}
				bcc.push_back(x);
				update();
			}
		}
		else
			low[x] = min(low[x], dfn[v]);
	}
}

int main() {
	while(1) {
		n = iread(); m = iread();
		if(n == 0 && m == 0) break;

		for(int i = 1; i <= n; i++) low[i] = dfn[i] = 0, head[i] = -1; cnt = 0;

		while(m--) {
			int u = iread(), v = iread();
			u++; v++;
			add(u, v); add(v, u);
		}

		ans1 = ans2 = clo = 0;
		for(int i = 1; i <= n; i++) if(!dfn[i]) tarjan(i, 0);

		printf("%d %d\n", ans1, ans2);
	}
	return 0;
}