Warm up

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=4612

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

Hint

题意

让你加一条边，使得这个图的桥数量最小

输出桥的数量

题解：

先缩点，然后求一个直径，然后输出树边数减去直径就好了

这样显然最小

代码

#include <bits/stdc++.h>

using namespace std;

const int maxn = 200000 + 500;

struct edge{

    int v , nxt;

}e[2005000];

int head[maxn] , tot , n , m , dfn[maxn] , low[maxn] , dfs_clock , bridge , dis[maxn] , vis[maxn] , Ftp , belong[maxn] ;

stack < int > sp;

vector < int > G[maxn];

void link(int u , int v){ e[tot].v=v,e[tot].nxt=head[u],head[u]=tot++;}

void dfs(int x , int pre){

	dfn[x] = low[x] = ++ dfs_clock; sp.push( x );

	for(int i = head[x] ; ~i ; i = e[i].nxt ){

		if( (i ^ 1) == pre ) continue;

		int v = e[i].v;

		if(!dfn[v]){

			dfs( v , i );

			low[x] = min( low[x] , low[ v ] );

			if(low[v] > dfn[x]) bridge ++ ;

		}else low[x]=min(low[x],dfn[v]);

	}

	if( low[x] == dfn[x] ){

		++ Ftp;

		while(1){

			int u = sp.top() ; sp.pop();

			belong[u] = Ftp;

			if( u == x ) break;

		}

	}

}

queue < int > Q;

int solve(){

	vis[1] = 1 ;

	Q.push( 1 );

	while(!Q.empty()){

		int s = Q.front() ; Q.pop();

		for(auto it : G[s]){

			if( vis[it] == 0 ){

				vis[it] = 1 ;

				dis[it] = dis[s] + 1;

				Q.push( it );

			}

		}

	}

	int s = -1 , index = 0;

	for( int it = 1 ; it <= Ftp ; ++ it ){

		if(dis[it] > s){

			s = dis[it];

			index = it;

		}

	}

	for(int i = 1 ; i <= Ftp ; ++ i) dis[i] = vis[i] = 0;

	Q.push( index );

	vis[index] = 1;

	while(!Q.empty()){

		int s = Q.front() ; Q.pop();

		for(auto it : G[s]){

			if( vis[it] == 0 ){

				vis[it] = 1 ;

				dis[it] = dis[s] + 1;

				Q.push( it );

			}

		}

	}

	s = 0;

	for( int it = 1 ; it <= Ftp ; ++ it ){

		if(dis[it] > s){

			s = dis[it];

		}

	}

	return s;

}

int main(int argc,char *argv[]){

	while(scanf("%d%d",&n,&m)){

		if( n == 0 && m == 0 ) break;

		for(int i = 1 ; i <= n ; ++ i) head[i] = -1 , dfn[i] = low[i] = vis[i] = dis[i] = 0 , G[i].clear();

		tot = dfs_clock = bridge = Ftp = 0 ;

		for(int i = 1 ; i <= m ; ++ i){

			int u , v ;

			scanf("%d%d",&u,&v);

			link( u , v );

			link( v , u );

		}

		for(int i = 1 ; i <= n ; ++ i) if(!dfn[i]) dfs( i , 1 << 30 );

		for(int i = 1 ; i <= n ; ++ i)

			for(int j = head[i] ; ~j ; j = e[j].nxt){

				int v = e[j].v;

				if(belong[i] == belong[v]) continue;

				G[belong[i]].push_back( belong[v] );

				G[belong[v]].push_back( belong[i] );

			}

		int maxv = solve();

		printf("%d\n" , bridge - maxv );

	}

	return 0;

}

巴特西

HDU 4612 Warm up tarjan 树的直径