POJ3177【边双连通分量缩点】

还不会双连通分量的朋友，请扣->这里<-

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题意：

No response.

思路：

在一个边双连通分量里面，所有的结点的low[ ]都是一样的哟。
所以可以缩点哟。
缩完点以后一定要证明哟。

//#include <bits/stdc++.h>#include<iostream>#include<cstdio>#include<cstring>#include<stack>#include<set>#include<map>#include<queue>#include<math.h>#include<algorithm>typedef long long LL;using namespace std;//#pragma comment(linker, "/STACK:102400000,102400000")const int INF=0x3f3f3f3f;const int N=5e3+10;struct Edge{    int to;    int next;    bool cut;} edge[N*4];int tol,head[N],n,m;void init(){    tol=0;    memset(head,-1,sizeof(head));}void add(int u,int v){    edge[tol].to=v;    edge[tol].next=head[u];    head[u]=tol++;}int low[N],dfn[N];int ind;void Tarjan(int u,int pre){    int v;    low[u]=dfn[u]=ind++;    for(int i=head[u]; ~i; i=edge[i].next)    {        v=edge[i].to;        if((i^1)==pre) continue;        if(!dfn[v]){            Tarjan(v,i);            low[u]=min(low[u],low[v]);        }        else            low[u]=min(low[u],dfn[v]);    }}int deg[N];void solve(){    memset(deg,0,sizeof(deg));    memset(dfn,0,sizeof(dfn));    ind=1;    Tarjan(1,-1);    for(int i=1;i<=n;i++)    {        for(int k=head[i];~k;k=edge[k].next)        {            int v=edge[k].to;            if(low[v] != low[i]) deg[low[i]]++;        }    }    int res=0;    for(int i=1;i<=n;i++)        if(deg[i]==1) res++;    printf("%d\n",(res+1)/2);}int main(){    int u,v;    scanf("%d%d",&n,&m);    init();    while(m--){        scanf("%d%d",&u,&v);        add(u,v);        add(v,u);    }    solve();    return 0;}