hdu4612

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

Total Submission(s): 3184 Accepted Submission(s): 720

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

题意：

给定一个联通图，问加入一条边后，最少还余下多少个割边

分析：先求强连通分量个数num，然后缩点形成一棵树，再求树的直径cnt，答案就是num-1-cnt；

程序：

#pragma comment(linker, "/STACK:10240000000000,10240000000000")

#include"stdio.h"

#include"string.h"

#include"stdlib.h"

#include"stack"

#include"iostream"

#define M 201009

#define inf 99999999

using namespace std;

stack<int>q;

struct st

{

     int u,v,w,next;

}edge[M*10];

int head[M],use[M],t,dis[M][3],in[M],index,num,belong[M],dfn[M],low[M];

void init()

{

     t=0;

     memset(head,-1,sizeof(head));

}

void add(int u,int v,int w)

{

     edge[t].u=u;

     edge[t].v=v;

     edge[t].w=w;

     edge[t].next=head[u];

     head[u]=t++;

}

void tarjan(int u,int id)

{

     dfn[u]=low[u]=++index;

     q.push(u);

     use[u]=1;

     int i;

     for(i=head[u];i!=-1;i=edge[i].next)

     {

          int v=edge[i].v;

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

          if(!dfn[v])

          {

               tarjan(v,i);

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

          }

          low[u]=min(low[u],dfn[v]);

     }

     if(dfn[u]==low[u])

     {

          int vv;

          num++;

          do

          {

               vv=q.top();

               q.pop();

               belong[vv]=num;

               use[vv]=0;

          }while(vv!=u);

     }

}

void dfs(int u)

{

     use[u]=1;

     for(int i=head[u];i!=-1;i=edge[i].next)

     {

          int v=edge[i].v;

          if(!use[v])

          {

               dfs(v);

               //更新最大值和次大值

               if(dis[u][0]<dis[v][0]+edge[i].w)

               {

                    int tt=dis[u][0];

                    dis[u][0]=dis[v][0]+edge[i].w;

                    dis[u][1]=tt;

               }

               else if(dis[u][1]<dis[v][0]+edge[i].w)

                    dis[u][1]=dis[v][0]+edge[i].w;

          }

     }

     if(in[u]==1&&u!=1)//注意

          dis[u][0]=dis[u][1]=0;

}

void solve(int n)

{

     index=num=0;

     memset(dfn,0,sizeof(dfn));

     memset(low,0,sizeof(low));

     memset(use,0,sizeof(use));

     memset(belong,0,sizeof(belong));

     tarjan(1,-1);

}

int uu[M],vv[M];

int main()

{

     int n,m,i;

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

     {

          init();

          while(m--)

          {

               int a,b;

               scanf("%d%d",&a,&b);

               add(a,b,1);

               add(b,a,1);

          }

          solve(n);

          int cnt=0;

          for(i=0;i<t;i+=2)

          {

               int u=edge[i].u;

               int v=edge[i].v;

               if(belong[u]!=belong[v])

               {

                    uu[cnt]=belong[u];

                    vv[cnt]=belong[v];

                    cnt++;

               }

          }

          init();

          memset(in,0,sizeof(in));

          memset(use,0,sizeof(use));

          memset(dis,0,sizeof(dis));

          for(i=0;i<cnt;i++)

          {

               //printf("%d %d\n",uu[i],vv[i]);

               add(uu[i],vv[i],1);

               add(vv[i],uu[i],1);

               in[uu[i]]++;

               in[vv[i]]++;

          }

          dfs(1);

          int ans=0;

          for(i=1;i<=num;i++)

          {

               if(ans<dis[i][0]+dis[i][1])

                    ans=dis[i][1]+dis[i][0];

          }

          printf("%d\n",num-1-ans);

     }

     return 0;

}

巴特西

tarjan算法求缩点+树形DP求直径

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