POJ 3177 Redundant Paths / 边双连通分量

和上一题一样 给你一张图 求最少加几条边可以使全图双连通 多了重边

多加了一个bool的数组 判断重边 重边只算一条

还是用了有向图的强连通分量 与有向图相比 多了 1.参数fa 父节点 2.如果子节点是其父节点 continue

#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;
const int maxn = 5010;

vector <int> G[maxn];
bool ok[maxn][maxn];
int pre[maxn];
int low[maxn];
int sccno[maxn];
int dfs_clock;
int scc_cnt;
stack <int> S;
int n, m;
int degree[maxn];
int Topo[maxn][maxn];
void dfs(int u, int fa)
{
	pre[u] = low[u] = ++dfs_clock;
	S.push(u);
	for(int i = 0; i < G[u].size(); i++)
	{
		int v = G[u][i];
		if(v == fa)
			continue;
		if(!pre[v])
		{
			dfs(v, u);
			low[u] = min(low[u], low[v]);
		}
		else
			low[u] = min(low[u], pre[v]);
	}
	if(low[u] == pre[u])
	{
		scc_cnt++;
		while(1)
		{
			int x = S.top();
			S.pop();
			sccno[x] = scc_cnt;
			if(x == u)
				break;
		}
	}
}
void find_scc()
{
	dfs_clock = scc_cnt = 0;
	memset(sccno, 0, sizeof(sccno));
	memset(pre, 0, sizeof(pre));
	for(int i = 1; i <= n; i++)
		if(!pre[i])
			dfs(i, -1);
}


int main()
{
	while(scanf("%d %d", &n, &m) != EOF)
	{
		for(int i = 1; i <= n; i++)
			G[i].clear();
		memset(ok, false, sizeof(ok));
		while(m--)
		{
			int u, v;
			scanf("%d %d", &u, &v);
			if(!ok[u][v])
			{
				G[u].push_back(v);
				G[v].push_back(u);
				ok[u][v] = true;
			}
		}
		find_scc();
		memset(degree, 0, sizeof(degree));
		for(int i = 1; i <= n; i++)
		{
			for(int j = 0; j < G[i].size(); j++)
			{
				int v = G[i][j];
				if(sccno[i] != sccno[v])
				{
					degree[sccno[i]]++;
					degree[sccno[v]]++;
				}
			}
		}
		int ans = 0;
		for(int i = 1; i <= scc_cnt; i++)
			if(degree[i] == 2)
				ans++;
		printf("%d\n", (ans + 1) / 2);
	}
	return 0;
}