[J]computer network tarjan边双联通分量+树的直径

https://odzkskevi.qnssl.com/b660f16d70db1969261cd8b11235ec99?v=1537580031

【2012-2013 ACM Central Region of Russia Quarterfinal Programming Contest】【J】computer network

题意：

n个点，m条边，构成一个无向图，现在让你再任意连接两个点，使得整个图的割边最少。

1 ≤ n ≤ 10 000; 1≤ m ≤ 100 000; 1 ≤ xi, yi ≤ n; xi ≠ yi.

题解：
tarjan求边双联通分量，缩点，形成一棵树，求树的直径，然后把直径的两个端点连起来就好。

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

// const int maxN=100;
// int a[maxN+10];
const int N=10010,M=100010;
int dfn[N],low[N],first[N],bfir[N],v[N],c[N],s[N],id[2];
int n,m,al,bl,sl,num,mx,cnt;
struct node{
    int x,y,next,tmp;
}a[2*M],b[2*M];

int minn(int x,int y){return x<y ? x:y;}

void ins(int x,int y)
{
    al++;a[al].tmp=0;
    a[al].x=x;a[al].y=y;a[al].next=first[x];first[x]=al;
}

void anins(int x,int y)
{
    bl++;b[bl].tmp=0;
    b[bl].x=x;b[bl].y=y;b[bl].next=bfir[x];bfir[x]=bl;
}

void tarjan(int x,int fa)
{
    dfn[x]=low[x]=++num;
    s[++sl]=x;
    for(int i=first[x];i;i=a[i].next)
    {
        if(a[i].tmp) continue;
        a[i%2==0 ? i-1:i+1].tmp=1;
        int y=a[i].y;
        if(!dfn[y])
        {
            tarjan(y,x);
            low[x]=minn(low[x],low[y]);
        }
        else if(!c[y]) low[x]=minn(low[x],dfn[y]);
    }
    if(dfn[x]==low[x])
    {
        cnt++;
        int z=0;
        while(1)
        {
            z=s[sl];sl--;
            v[z]=0;
            c[z]=cnt;
            if(z==x) break;
        }
    }
}

void dfs(int x,int fa,int dep,int tmp)
{
    if(dep>mx) mx=dep,id[tmp]=x;
    for(int i=bfir[x];i;i=b[i].next)
    {
        int y=b[i].y;
        if(y==fa) continue;
        dfs(y,x,dep+1,tmp);
    }
}

int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    scanf("%d%d",&n,&m);
    al=0;sl=0;bl=0;num=0;cnt=0;
    memset(bfir,0,sizeof(bfir));
    memset(first,0,sizeof(first));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(c,0,sizeof(c));
    for(int i=1;i<=m;i++)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        ins(x,y);ins(y,x);
    }
    for(int i=1;i<=n;i++)
    {
        if(!dfn[i]) num=0,tarjan(i,0);
    }
    // for(int i=1;i<=n;i++) printf("c [ %d ] = %d
",i,c[i]);
    for(int i=1;i<=2*m;i++)
    {
        int x=a[i].x,y=a[i].y;
        if(c[y]==c[x]) continue;
        anins(c[x],c[y]);
        // printf("%d -- > %d
",c[x],c[y]);
    }
    mx=0,dfs(1,0,0,0);
    mx=0,dfs(id[0],0,0,1);
    for(int i=1;i<=n;i++)
        if(c[i]==id[0]) {printf("%d ",i);break;}
    for(int i=1;i<=n;i++)
        if(c[i]==id[1]) {printf("%d
",i);break;}
    // printf("%d %d
",id[0],id[1]);
    return 0;
}