POJ——T3352 Road Construction

http://poj.org/problem?id=3352

vis表示访问的次序 low的值相同的点在同一连通分量

#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;

const int N(2017);
int n,m,u,v,ans;
int head[N],sumedge;
struct Edge
{
    int to,next;
    Edge(int to=0,int next=0) :
        to(to),next(next) {}
}edge[N<<1];

void ins(int from,int to)
{
    edge[++sumedge]=Edge(to,head[from]);
    head[from]=sumedge;
}

int tim,dfn[N],low[N],vis[N],du[N];
int sumcol,col[N],colvis[N];

void DFS(int now,int pre)
{
    low[now]=dfn[now]=++tim;
    vis[now]=1;
    for(int i=head[now];i;i=edge[i].next)
    {
        int go=edge[i].to;
        if(go==pre) continue;
        if(!vis[go])
        {
            DFS(go,now);
            low[now]=min(low[now],low[go]);
        } else if(vis[go]==1)
            low[now]=min(low[now],dfn[go]);
    }
    vis[now]=2;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(;m;m--)
    {
        scanf("%d%d",&u,&v);
        ins(u,v); ins(v,u);
    }
    for(int i=1;i<=n;i++)
        if(!vis[i]) DFS(i,i);
    for(u=1;u<=n;u++)
        for(int j=head[u];j;j=edge[j].next)
        {
            v=edge[j].to;
            if(low[u]!=low[v])
                du[low[u]]++;
        }
    for(int i=1;i<=n;i++)
        if(du[i]==1) ans++;  
    printf("%d
",(ans+1)>>1);
    return 0;
}