无向图边双联通分量+缩点

无向图边双联通分量+缩点

先求桥,再遍历一遍图,把桥去掉


#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <deque>
#include <map>
#include <iostream>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0);
const double e = exp(1);
//const int MAXN =2e5+10;
const LL N = 1000000007;

struct edge
{
	int id;
	int flag;
	int from;
	int to;
	int next;
} edge[200009], edge3[200009];
int head[200009];
int head3[200009]; //描述新图中结点之间的关系

int dfn[100009];
int low[100009];
int cnt = 1;
int cnt3 = 0; //用于构建缩点后新图的链式前向星

int New[100009];	 //第i个双联通分量所含有的结点个数
int captain[100009]; //点i所在的边双连通分量

int father[100009];
int vis[100009];
int deep[100009];
int sum;

void tarjan(int u, int id)  //需要考虑去重边的问题
{
	int i;
	low[u] = dfn[u] = cnt++;
	for (i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if(id == edge[i].id)     //判断是不是同一条边,若id相同则为同一条
			continue;
		if (!dfn[v])
		{
			tarjan(v, edge[i].id);
			low[u] = min(low[u], low[v]);
			if (low[v] > dfn[u])
			{
				edge[i].flag = edge[i ^ 1].flag = 1; // 标记割边,在寻找边双连通分量时忽略掉该边

			//	printf("%d ---> %d\n", edge[i].from, edge[i].to);
			}
		}
		else
		{
			low[u] = min(low[u], dfn[v]);
		}
	}
}

void seek_doubleEdge(int u, int cnt2)
{
	int i;
	dfn[u] = cnt2;
	New[cnt2]++;
	captain[u] = cnt2;
	for (i = head[u]; i != -1; i = edge[i].next)
	{
		int v = edge[i].to;
		if (edge[i].flag)
		{

			if (captain[v])
			{
				int a = captain[u];
				int b = captain[v];

				edge3[cnt3].to = b;
				edge3[cnt3].next = head3[a];
				head3[a] = cnt3++;

				edge3[cnt3].to = a;
				edge3[cnt3].next = head3[b];
				head3[b] = cnt3++;

				//	cout << u << " " << v << " " << a << " "<< b << " ** " << endl;
			}
			continue;
		}
		else
		{
			if (!dfn[v])
			{
				seek_doubleEdge(v, cnt2);
			}
		}
	}
}

void init_lca(int u, int fa, int d)
{
	int i;
	int num = 0;
	deep[u] = d;
	father[u] = fa;
	//	printf("%d  %d  ?? \n", u, deep[u]);
	for (i = head3[u]; i != -1; i = edge3[i].next)
	{
		int v = edge3[i].to;
		if (!deep[v])
		{
			//printf("%d %d %d  (^_^)\n", v, u, d);
			num++;
			init_lca(v, u, d + 1);
		}
	}

	if(num == 0 || (u == 1 && num == 1))
	{
		//cout << u << endl;
		sum++;
	}
}


int main()
{
	int n, m, i;
	int cnt1, a, b, q, nn = 0;
	while (scanf("%d%d", &n, &m) != EOF)
	{
		if (n == 0 && m == 0)
			break;
		nn++;
		cnt1 = 0;
		cnt = 1;
		memset(head, -1, sizeof(head));
		memset(dfn, 0, sizeof(dfn));
		memset(low, 0, sizeof(low));
		memset(captain, 0, sizeof(captain));
		while (m--)
		{
			scanf("%d%d", &a, &b);

			edge[cnt1].id = cnt1;          //[1]
			edge[cnt1].flag = 0;
			edge[cnt1].from = a;
			edge[cnt1].to = b;
			edge[cnt1].next = head[a];
			head[a] = cnt1++;

			edge[cnt1].id = cnt1 - 1;      //[2],同[1]处注释一起为一条无向边的来回方向都标记同一个id.
			edge[cnt1].flag = 0;
			edge[cnt1].from = b;
			edge[cnt1].to = a;
			edge[cnt1].next = head[b];
			head[b] = cnt1++;
		}
		for (i = 1; i <= n; i++) //双连通分量去割边
		{
			if (!dfn[i])
			{
				tarjan(i, -1);
			}
		}

		memset(dfn, 0, sizeof(dfn));
		memset(head3, -1, sizeof(head3));
		memset(vis, 0, sizeof(vis));
		memset(deep, 0, sizeof(deep));

		int cnt2 = 0;
		for (int i = 1; i <= n; i++) //标记双连通分量(缩点)
		{
			if (!dfn[i])
			{
				cnt2++;
				seek_doubleEdge(i, cnt2);
			}
		}
/*
		cout << " ** " << cnt2 << endl; // 边双连通分量的个数
		for (int i = 1; i <= n; i++)
		{
			printf("* %d   %d\n", i, captain[i]);
		}

		
		for (int i = 1; i <= cnt2; i++) //输出缩点后的新图
		{
			for (int j = head3[i]; j != -1; j = edge3[j].next)
			{
				cout << i << " " << edge3[j].to << endl;
			}
		}
*/
		sum = 0;
		init_lca(1, -1, 1);
	//	cout << "sum: " << sum << endl;
		if(sum == 1)
			printf("0\n");
		else
			printf("%d\n", (sum + 1) / 2);

/*		for (int i = 1; i <= cnt2; i++)
		{
			cout << i << " " << deep[i] << "  ?? " << endl;
		}
*/
		
	}
	return 0;
}
posted @ 2019-04-26 22:33  Daybreaking  阅读(103)  评论(0编辑  收藏  举报