HDU 3849 By Recognizing These Guys, We Find Social Networks Useful

题目链接

题意：说白了就是求一个无向图的桥

思路：字符串hash掉，然后双连通。要注意特判一下假设不是一个连通块。那么答案是0

代码：

#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <map>
using namespace std;

const int N = 10005;
const int M = 200005;
int t, n, m;
map<string, int> hash;
char A[20], B[20];

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 en, first[N], next[M], hn;
char name[N][20];

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

int get(char *str) {
	if (!hash.count(str)) {
		strcpy(name[hn], str);
		hash[str] = hn++;
	}
	return hash[str];
}

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

int pre[N], dfn[N], dfs_clock, ans = 0;

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

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

int parent[N];

int find(int x) {
	return parent[x] == x ?

x : parent[x] = find(parent[x]);
}
int main() {
	scanf("%d", &t);
	while (t--) {
		init();
		scanf("%d%d", &n, &m);
		for (int i = 0; i < n; i++) parent[i] = i;
		int cnt = n;
		for (int i = 0; i < m; i++) {
			scanf("%s%s", A, B);
			int u = get(A), v = get(B);
			add_edge(u, v, i);
			add_edge(v, u, i);
			int pu = find(u), pv = find(v);
			if (pu != pv) {
				cnt--;
				parent[pu] = pv;
			}
		}
		if (cnt > 1) {
			printf("0\n");
			continue;
		}
		ans = 0;
		find_cut();
		printf("%d\n", ans);
		for (int i = 0; i < en; i += 2)
			if (edge[i].iscut)
				printf("%s %s\n", name[edge[i].u], name[edge[i].v]);
	}
	return 0;
}