hdoj 4612 Warm up 【无向图边双连通求桥数 缩点后求树的直径 桥数-树的直径】

Warm up

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4569    Accepted Submission(s): 1031

Problem Description

　　N planets are connected by M bidirectional channels that allow instant transportation. It's always possible to travel between any two planets through these channels.
　　If we can isolate some planets from others by breaking only one channel , the channel is called a bridge of the transportation system.
People don't like to be isolated. So they ask what's the minimal number of bridges they can have if they decide to build a new channel.
　　Note that there could be more than one channel between two planets.

 

Input

　　The input contains multiple cases.
　　Each case starts with two positive integers N and M , indicating the number of planets and the number of channels.
　　(2<=N<=200000, 1<=M<=1000000)
　　Next M lines each contains two positive integers A and B, indicating a channel between planet A and B in the system. Planets are numbered by 1..N.
　　A line with two integers '0' terminates the input.

 

Output

　　For each case, output the minimal number of bridges after building a new channel in a line.

 

Sample Input

4 4
1 2
1 3
1 4
2 3
0 0 

 

Sample Output

0

 

题意：给出n个点，m条边。 问：加上一条边使得桥数最少，让你求出该桥数。

 

求桥数，缩点后求树的直径，然后桥数减树的直径。

 

恶心死了：重边那里就因为写成u == fa WA 8次，最后才发现是v == fa。无语

 

#pragma comment(linker, "/STACK:1024000000,1024000000")//预处理栈大小 
#include <cstdio>
#include <cstring>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#define MAXN 200000+10 
#define MAXM 2000000+10 
#define INF 100000000
using namespace std;
struct Edge
{
	int from, to, next;
	bool cut;//是否为桥 
}edge[MAXM];
int head[MAXN], edgenum; 
int dfs_clock;//dfs次数 时间戳的 应用 
int low[MAXN], dfn[MAXN];
int ebcno[MAXN], ebc_cnt;//ebcno[i]表示i属于哪个EBC    ebc_cnt是EBC计数器 
int bridge;//桥数 
int n, m;//点数 边数
stack<int> S;//存储当前bcc的所有点
bool Instack[MAXN];//标记该点是否在栈里面 
vector<int> G[MAXN];//存储缩点后新图 
int dist[MAXN];//对新图存储源点的 最长路 
bool vis[MAXN];//标记是否查询过 
int node;//最长路S - T端点 
int ans;//树的直径 
void init()
{
	edgenum = 0;
	memset(head, -1, sizeof(head));
} 
void addEdge(int u, int v)
{
	Edge E = {u, v, head[u], 0};
	edge[edgenum] = E;
	head[u] = edgenum++; 
}
void getMap()//建图 
{
	int a, b;
	while(m--)
	{
		scanf("%d%d", &a, &b);
		addEdge(a, b);
		addEdge(b, a);
	} 
}
void tarjan(int u, int fa)
{
	int v;
	low[u] = dfn[u] = ++dfs_clock;//时间戳 
	S.push(u);
	Instack[u] = true;//进栈
	int have = 1;
	for(int i = head[u]; i != -1; i = edge[i].next)
	{
		v = edge[i].to;
		if(have && v == fa)//重边好恶心 因为这里WA  8次 
		{
			have = 0;
			continue;
		}
		if(!dfn[v])
		{
			tarjan(v, u);
			low[u] = min(low[u], low[v]);
			if(low[v] > dfn[u])//桥 
			{
				bridge++;
				edge[i].cut = edge[i^1].cut = true;
			}
		}
		else if(Instack[v])//已搜索过 且 在栈里面 
		low[u] = min(low[u], dfn[v]);
	} 
	if(low[u] == dfn[u]) 
	{
		ebc_cnt++;
		for(;;)//弹出属于同一个EBC的点 
		{
			v = S.top(); S.pop();
	     	ebcno[v] = ebc_cnt;//属于当前EBC 
			Instack[v] = false;
			if(v == u) break; 
		} 
	}
}
void suodian()//缩点  建新图 
{
	for(int i = 1; i <= ebc_cnt; i++) G[i].clear();//初始化 
	for(int i = 0; i < edgenum; i+=2)//跳过重边 
	{
		int u = ebcno[edge[i].from];
		int v = ebcno[edge[i].to];
		if(u != v)//不在同一个EBC
		G[u].push_back(v), G[v].push_back(u); 
	}
}
void find_cut(int l, int r)
{
	memset(low, 0, sizeof(low));
	memset(dfn, 0, sizeof(dfn));
	memset(Instack, false, sizeof(Instack));
	memset(ebcno, 0, sizeof(ebcno));
	dfs_clock = ebc_cnt = bridge = 0;
	for(int i = l; i <= r; i++)
	if(!dfn[i]) tarjan(i, -1);
} 
void BFS(int start)
{
	queue<int> Q;
	memset(dist, 0, sizeof(dist));
	memset(vis, 0, sizeof(vis));
	node = start;
	ans = 0;
	vis[start] = 1;
	Q.push(start);
	while(!Q.empty())
	{
		int u = Q.front();
		Q.pop();
		for(int i = 0; i < G[u].size(); i++)
		{
			int v = G[u][i];
			if(!vis[v])
			{
				dist[v] = dist[u] + 1;
				if(dist[v] > ans)
				{
					ans = dist[v];
					node = v;
				} 
				vis[v] = 1; 
				Q.push(v);
			} 
		}
	}
}
int main()
{
	while(scanf("%d%d", &n, &m), n||m)
	{
		init();
		getMap();
		find_cut(1, n);
		suodian();//缩点 
		BFS(1);//找最长路 S-T 端点 
		BFS(node);//最长路 
		printf("%d\n", bridge - ans);//桥数 - 树的直径 
	} 
	return 0;
}