HDU4612 Warm up 边双联通分量+树的直径-优快云博客

本文探讨了在连接多个行星的运输系统中，通过构建新通道来最小化桥梁数量的问题。通过使用深度优先搜索（DFS）算法两次遍历，解决了如何找到树的直径并据此计算最少桥梁数。

模板题

桥的数量-树的直径

输的直径两次DFS即可

Warm up

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 2506 Accepted Submission(s): 586

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

Sample Output

#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<vector>
#include<queue>

using namespace std;

#define MAXN 200100
#define MAXM 9200100

struct node
{
    int from,to,next;
    bool vis;
}edge[MAXM];

int head[MAXN],en;
int n,m,dfnCnt,dfn[MAXN],low[MAXN];
bool jud[MAXM],vis[MAXN];
int stack[MAXN],top,color[MAXN],col;
int bridgeU[MAXN],bridgeV[MAXN],bri;

void add(int a,int b)
{
    edge[en].to=b;
    edge[en].from=a;
    edge[en].next=head[a];
    edge[en].vis=0;
    head[a]=en++;
    edge[en].to=a;
    edge[en].from=b;
    edge[en].next=head[b];
    edge[en].vis=0;
    head[b]=en++;
}

void dfs(int u)
{
    vis[u]=1;
    dfn[u]=low[u]=++dfnCnt;
    stack[++top]=u;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].to;
        if(edge[i].vis) continue;
        edge[i].vis=edge[i^1].vis=1;
        if(!vis[v])
        {
            dfs(v);
            low[u]=min(low[u],low[v]);
            if(low[v]>dfn[u])
            {
                int ID=i/2+1;
                jud[ID]=1;
                bridgeU[bri]=u;
                bridgeV[bri++]=v;
            }
        }
        else
           low[u]=min(low[u],dfn[v]);
    }

    if(low[u]==dfn[u])
    {
        col++;
        do
        {
            color[stack[top]]=col;
        }
        while(stack[top--]!=u);
    }
}

void tarjan()
{
    bri=0;col=0;
    memset(vis,0,sizeof(vis));
    memset(jud,0,sizeof(jud));
    memset(color,-1,sizeof(color));
    for(int i=1;i<=n;i++)
    {
        dfnCnt=0;top=0;
        if(!vis[i])
        {
            dfs(i);
        }
    }
}


int dist[MAXN];
int head2[MAXN];

void add2(int a,int b)
{
    edge[en].to=b;
    edge[en].from=a;
    edge[en].next=head2[a];
    head2[a]=en++;
    edge[en].to=a;
    edge[en].from=b;
    edge[en].next=head2[b];
    head2[b]=en++;
}


void DFS(int len,int fa,int now)
{
    dist[now]=len;
    for(int i=head2[now];i!=-1;i=edge[i].next)
    {
        int v=edge[i].to;
        if(v!=fa) DFS(len+1,now,v);
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0 && m==0) break;
        memset(head,-1,sizeof(head));en=0;
        for(int i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            add(u,v);
        }
        tarjan();
        if(bri==0)
        {
            printf("0\n");
            continue;
        }
        memset(head2,-1,sizeof(head2));
        for(int i=0;i<bri;i++)
        {
            int u=bridgeU[i];
            int v=bridgeV[i];
            if(color[u]==color[v]) continue;
            add2(color[u],color[v]);
        }
        DFS(0,-1,1);
        int maxn=0;int kk;
        for(int i=1;i<=col;i++)
        {
            if(dist[i]>maxn)
            {
                maxn=dist[i];
                kk=i;
            }
        }
        DFS(0,-1,kk);
        maxn=0;
        for(int i=1;i<=col;i++)
        {
            if(dist[i]>maxn)
            {
                maxn=dist[i];
            }
        }
        printf("%d\n",bri-maxn);
    }
    return 0;
}