POJ 3177 Redundant Paths

POJ 3352 Road Construction

题目链接

题意：两题一样的，一份代码能交，给定一个连通无向图，问加几条边能使得图变成一个双连通图

思路：先求双连通，缩点后，计算入度为1的个数，然后（个数 + 1） / 2 就是答案（这题由于是只有一个连通块所以可以这么搞，如果有多个，就不能这样搞了）

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 1005;
const int M = 20005;

int n, m;
struct Edge {
	int u, v, id;
	bool iscut;
	Edge() {}
	Edge(int u, int v, int id) {
		this->u = u;
		this->v = v;
		this->id = id;
		this->iscut = false;
	}
} edge[M];

int first[N], next[M], en;

void add_edge(int u, int v, int id) {
	edge[en] = Edge(u, v, id);
	next[en] = first[u];
	first[u] = en++;
}

void init() {
	en = 0;
	memset(first, -1, sizeof(first));
}

int pre[N], dfn[N], dfs_clock, bccno[N], bccn;

void dfs_cut(int u, int id) {
	pre[u] = dfn[u] = ++dfs_clock;
	for (int i = first[u]; i + 1; i = next[i]) {
		if (edge[i].id == id) continue;
		int v = edge[i].v;
		if (!pre[v]) {
			dfs_cut(v, edge[i].id);
			dfn[u] = min(dfn[u], dfn[v]);
			if (dfn[v] > pre[u])
				edge[i].iscut = edge[i^1].iscut = true;
		} else dfn[u] = min(dfn[u], pre[v]);
	}
}

void find_cut() {
	dfs_clock = 0;
	memset(pre, 0, sizeof(pre));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs_cut(i, -1);
}

void dfs_bcc(int u) {
	bccno[u] = bccn;
	for (int i = first[u]; i + 1; i = next[i]) {
		if (edge[i].iscut) continue;
		int v = edge[i].v;
		if (bccno[v]) continue;
		dfs_bcc(v);
	}
}

void find_bcc() {
	bccn = 0;
	memset(bccno, 0, sizeof(bccno));
	for (int i = 1; i <= n; i++) {
		if (!bccno[i]) {
			bccn++;
			dfs_bcc(i);
		}
	}
}

int du[N];

int main() {
	while (~scanf("%d%d", &n, &m)) {
		int u, v;
		init();
		for (int i = 0; i < m; i++) {
			scanf("%d%d", &u, &v);
			add_edge(u, v, i);
			add_edge(v, u, i);
		}
		find_cut();
		find_bcc();
		memset(du, 0, sizeof(du));
		for (int i = 0; i < en; i += 2) {
			if (!edge[i].iscut) continue;
			int u = bccno[edge[i].u], v = bccno[edge[i].v];
			if (u == v) continue;
			du[u]++; du[v]++;
		}
		int cnt = 0;
		for (int i = 1; i <= bccn; i++)
			if (du[i] == 1) cnt++;
		printf("%d\n", (cnt + 1) / 2);
	}
	return 0;
}