【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  2.718281828459
typedef 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;
}