【POJ2186】Popular Cows(tarjan+缩点)

记录一个菜逼的成长。。

第一道强连通图的题。。
tarjan算法+缩点 模板题
算法伪代码如下

tarjan(u)
{
DFN[u]=Low[u]=++Index // 为节点u设定次序编号和Low初值
Stack.push(u) // 将节点u压入栈中
for each (u, v) in E // 枚举每一条边
if (v is not visted) // 如果节点v未被访问过
tarjan(v) // 继续向下找
Low[u] = min(Low[u], Low[v])
else if (v in S) // 如果节点v还在栈内
Low[u] = min(Low[u], DFN[v])
if (DFN[u] == Low[u]) // 如果节点u是强连通分量的根
repeat
v = S.pop // 将v退栈，为该强连通分量中一个顶点
print v
until (u== v)
}

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <stack>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define ALL(v) (v).begin(),(v).end()#define cl(a,b) memset(a,b,sizeof(a))#define bp __builtin_popcount#define pb push_back#define mp make_pair#define fin freopen("D://in.txt","r",stdin)#define fout freopen("D://out.txt","w",stdout)#define lson t<<1,l,mid#define rson t<<1|1,mid+1,r#define seglen (node[t].r-node[t].l+1)#define pi 3.1415926#define exp  2.718281828459typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;typedef vector<PII> VPII;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;template <typename T>inline void read(T &x){    T ans=0;    char last=' ',ch=getchar();    while(ch<'0' || ch>'9')last=ch,ch=getchar();    while(ch>='0' && ch<='9')ans=ans*10+ch-'0',ch=getchar();    if(last=='-')ans=-ans;    x = ans;}/******************head***********************/const int maxn = 50000 + 10;int head[maxn],cnt;struct Edge{    int next,to;}edge[maxn];void add(int u,int v){    edge[cnt].to = v;    edge[cnt].next = head[u];    head[u] = cnt++;}int instack[maxn],dfn[maxn],low[maxn],ind,num,belong[maxn],setnum[maxn];stack<int>sta;int d[maxn];///出度void tarjan(int u){    ///dfn(u)为时间戳，即dfs序    ///Low(u)为u或u的子树能够追溯到的最早的栈中节点的次序号    dfn[u] = low[u] = ++ind;    instack[u] = 1;    sta.push(u);    for( int i = head[u]; ~i; i = edge[i].next ){        int v = edge[i].to;        if(!dfn[v]){            tarjan(v);            low[u] = min(low[u],low[v]);        }        else if(instack[v]){            low[u] = min(low[u],dfn[v]);        }    }    if(dfn[u] == low[u]){        num++;        int v;        do{            v = sta.top();            sta.pop();            instack[v] = 0;            belong[v] = num;///点v属于第num个强连通分量            setnum[num]++;///每个强连通分量的数量        }while(u != v);    }}void init(){    while(!sta.empty())sta.pop();    cl(head,-1),cnt = 0;    cl(instack,0),cl(dfn,0),cl(low,0);    ind = 0;    num = 0;    cl(belong,0);    cl(setnum,0);    cl(d,0);}int main(){    //fin,fout;    int n,m;    while(~scanf("%d%d",&n,&m)){        init();        int u,v;        for( int i = 1; i <= m; i++ ){            scanf("%d%d",&u,&v);            add(u,v);        }        for( int i = 1; i <= n; i++ ){            if(!dfn[i]){                tarjan(i);            }        }       for( int i = 1; i <= n; i++ ){            for(int j = head[i]; ~j; j = edge[j].next ){                if(belong[i] != belong[edge[j].to]){                    d[belong[i]]++;                }            }        }        int cnt = 0,res,ans = 0;        for( int i = 1; i <= num; i++ ){            if(!d[i]){                cnt++;                res = i;            }        }        printf("%d\n",cnt == 1 ? setnum[res] : 0);    }    return 0;}