poj 3180 The Cow Prom（强连通分量 Tarjan）

来源：互联网发布：百度人工智能研究院编辑：程序博客网时间：2024/04/29 06:35

题意：有N个牛，围在一水池边，它们用绳子互相绑着（有方向的）。如果绳子的方向一致，它们就能顺时针转，问有多少组牛可以跳舞。

思路：简单有向图的强连通分量。求出强连通分量，且强连通分量里的点数大于等于2的块就能跳舞。

//836K    79MS#include#include#includeusing namespace std;const int VM = 10005;const int EM = 50005;struct Edge{    int to,nxt;}edge[EM];int head[VM],vis[VM],dfn[VM],low[VM];int stack[VM+10],belong[VM];int scc,cnt,top,ep;void addedge (int cu,int cv){    edge[ep].to = cv;    edge[ep].nxt = head[cu];    head[cu] = ep++;}int min (int a ,int b){    return a > b ? b : a;}void Tarjan(int u){    dfn[u] = low[u] = ++cnt;    vis[u] = 1;    stack[top++] = u;    int v;    for (int i = head[u];i != -1;i = edge[i].nxt)    {        v = edge[i].to;        if (!dfn[v])        {            Tarjan(v);            low[u] = min(low[u],low[v]);        }        else if (vis[v]) low[u] = min(low[u],dfn[v]);    }    if (dfn[u] == low[u])    {        ++scc;        do{            v = stack[--top];            vis[v] = 0;            belong[v] = scc;        }while (u != v);    }}void solve(int n){    memset (vis,0,sizeof(vis));    memset (dfn,0,sizeof(dfn));    scc = cnt = top = 0;    int u,v;    for (u = 1;u <= n;u ++)        if (!dfn[u])            Tarjan(u);    sort (belong+1,belong+n+1);    belong[n+1] = -1;   // for (int i = 0;i <= n;i ++)   //     printf ("%d ",belong[i]);    int pre = belong[1];    cnt = 0;    int ans = 0;    for (int i = 1;i <= n+1;i ++)    {        if (pre == belong[i])            ++cnt;        else {            pre = belong[i];            if (cnt >= 2) ans ++;            cnt = 1;        }    }    printf ("%d\n",ans);}int main (){    #ifdef LOCAL        freopen("in.txt","r",stdin);    #endif    int n,m,u,v;    while (~scanf ("%d%d",&n,&m))    {        memset (head,-1,sizeof(head));        ep = 0;        while (m --)        {            scanf ("%d%d",&u,&v);            addedge (u,v);        }        solve(n);    }    return 0;}