[HAOI 2006]受欢迎的牛

Description

　　每一头牛的愿望就是变成一头最受欢迎的牛。现在有N头牛，给你M对整数(A,B)，表示牛A认为牛B受欢迎。这种关系是具有传递性的，如果A认为B受欢迎，B认为C受欢迎，那么牛A也认为牛C受欢迎。你的任务是求出有多少头牛被所有的牛认为是受欢迎的。

Input

　　第一行两个数N,M。接下来M行，每行两个数A,B，意思是A认为B是受欢迎的（给出的信息有可能重复，即有可

能出现多个A,B）

Output

　　一个数，即有多少头牛被所有的牛认为是受欢迎的。

Sample Input

3 3
1 2
2 1
2 3

Sample Output

HINT

100%的数据N<=10000,M<=50000

题解

好久没打$tarjan$了，码在这当模板存着。

 //It is made by Awson on 2017.9.24

 #include <set>

 #include <map>

 #include <cmath>

 #include <ctime>

 #include <queue>

 #include <stack>

 #include <vector>

 #include <string>

 #include <cstdio>

 #include <cstdlib>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 #define LL long long

 #define Min(a, b) ((a) < (b) ? (a) : (b))

 #define Max(a, b) ((a) > (b) ? (a) : (b))

 using namespace std;

 const int N = ;

 const int M = ;

 int Read() {

     char ch = getchar();

     int sum = ;

     while (ch < '' || ch > '') ch = getchar();

     while (ch >= '' && ch <= '') sum = (sum<<)+(sum<<)+ch-, ch = getchar();

     return sum;

 }

 int n, m, u, v;

 struct tt {

     int from, to, next;

 }edge[M+];

 int path[N+], top;

 int dfn[N+], low[N+], t;

 bool vis[N+];

 int S[N+], topS;

 int sccno[N+], sccnum;

 int in[N+], tot[N+];

 void add(int u, int v) {

     edge[++top].to = v;

     edge[top].from = u;

     edge[top].next = path[u];

     path[u] = top;

 }

 void tarjan(int r) {

     dfn[r] = low[r] = ++t;

     vis[r] = ;

     S[topS++] = r;

     for (int i = path[r]; i; i = edge[i].next) {

         if (!dfn[edge[i].to]) {

             tarjan(edge[i].to);

             low[r] = Min(low[edge[i].to], low[r]);

         }

         else if (vis[edge[i].to])

             low[r] = Min(low[r], dfn[edge[i].to]);

     }

     if (dfn[r] == low[r]) {

         sccnum++;

         while (topS >  && S[topS] != r) {

             vis[S[--topS]] = ;

             sccno[S[topS]] = sccnum;

             tot[sccnum]++;

         }

     }

 }

 void work() {

     n = Read(), m = Read();

     for (int i = ; i <= m; i++) {

         u = Read(), v = Read();

         add(u, v);

     }

     for(int i = ; i <= n; i++)

         if(!dfn[i]) tarjan(i);

     for (int i = ; i <= m; i++)

         if (sccno[edge[i].to] != sccno[edge[i].from])

                in[sccno[edge[i].from]]++;

     int ans = , cntt = ;

     for (int i = ; i <= sccnum; i++)

         if (!in[i]) cntt++, ans=i;

     if (cntt == ) printf("%d\n", tot[ans]);

     else printf("0\n");

 }

 int main() {

     work();

     return ;

 }

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