题目描述
The cows are having a picnic! Each of Farmer John‘s K (1 ≤ K ≤ 100) cows is grazing in one of N (1 ≤ N ≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected by M (1 ≤ M ≤ 10,000) one-way paths (no path connects a pasture to itself).
The cows want to gather in the same pasture for their picnic, but (because of the one-way paths) some cows may only be able to get to some pastures. Help the cows out by figuring out how many pastures are reachable by all cows, and hence are possible picnic locations.
K(1≤K≤100)只奶牛分散在N(1≤N≤1000)个牧场．现在她们要集中起来进餐．牧场之间有M(1≤M≤10000)条有向路连接，而且不存在起点和终点相同的有向路．她们进餐的地点必须是所有奶牛都可到达的地方．那么，有多少这样的牧场呢？
输入输出格式
输入格式：
Line 1: Three space-separated integers, respectively: K, N, and M
Lines 2..K+1: Line i+1 contains a single integer (1..N) which is the number of the pasture in which cow i is grazing.
Lines K+2..M+K+1: Each line contains two space-separated integers,
respectively A and B (both 1..N and A != B), representing a one-way path
from pasture A to pasture B.
输出格式：
Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.
输入输出样例
输入样例#1：
复制
2 4 4 2 3 1 2 1 4 2 3 3 4
输出样例#1： 复制
2
说明
The cows can meet in pastures 3 or 4.
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<time.h>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 260005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 98765431;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline int rd() {
int x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n, m, K;
vector<int>vc[1002];
int cow[102];
int num[1002];
int vis[1002];
void dfs(int u) {
int siz = vc[u].size();
num[u]++; vis[u] = 1;
for (int i = 0; i < siz; i++) {
int v = vc[u][i];
if (!vis[v])
dfs(v);
}
return;
}
int main()
{
// ios::sync_with_stdio(0);
K = rd(); n = rd(); m = rd();
for (int i = 1; i <= K; i++)cow[i] = rd();
for (int i = 1; i <= m; i++) {
int u = rd(), v = rd();
vc[u].push_back(v);
}
for (int i = 1; i <= K; i++) {
ms(vis);
dfs(cow[i]);
}
int ans = 0;
for (int i = 1; i <= n; i++) {
// cout << i << ‘ ‘ << num[i] << endl;
if (num[i] == K)ans++;
}
cout << ans << endl;
return 0;
}
原文地址：https://www.cnblogs.com/zxyqzy/p/10371617.html