1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐
Time Limit: 5 Sec Memory Limit: 64 MB Submit: 781 Solved: 483 [Submit][Status][Discuss]
Description
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？
Input
* 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.
第1行输入K，N，M.接下来K行，每行一个整数表示一只奶牛所在的牧场编号．接下来M行，每行两个整数，表示一条有向路的起点和终点
Output
* Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.
所有奶牛都可到达的牧场个数
Sample Input
2 4 4
2
3
1 2
1 4
2 3
3 4
INPUT DETAILS:
4<--3
^ ^
| |
| |
1-->2
The pastures are laid out as shown above, with cows in pastures 2 and 3.
Sample Output
2
牧场3，4是这样的牧场．
题目说的是没有自环，我以为说的是DAG。。。WA了半天
直接暴力bfs就行了
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
char buf[10000000], *ptr = buf - 1;
inline int readint(){
int f = 1, n = 0;
char ch = *++ptr;
while(ch < ‘0‘ || ch > ‘9‘){
if(ch == ‘-‘) f = -1;
ch = *++ptr;
}
while(ch <= ‘9‘ && ch >= ‘0‘){
n = (n << 1) + (n << 3) + ch - ‘0‘;
ch = *++ptr;
}
return f * n;
}
const int maxn = 1000 + 10, maxm = 10000 + 10;
struct Edge{
int to, next;
Edge(){}
Edge(int _t, int _n): to(_t), next(_n){}
}e[maxm];
int fir[maxn] = {0}, cnt = 0;
inline void add(int u, int v){
e[++cnt] = Edge(v, fir[u]); fir[u] = cnt;
}
int k, n, m, ans = 0;
int w[maxn], f[maxn] = {0}, vis[maxn] = {0};
int q[maxn], h, t;
void work(){
for(int u, v, i = 1; i <= k; i++){
h = t = 0;
q[t++] = w[i];
vis[w[i]] = i;
while(h != t){
u = q[h++];
f[u]++;
for(int j = fir[u]; j; j = e[j].next){
v = e[j].to;
if(vis[v] != i){
q[t++] = v;
vis[v] = i;
}
}
}
}
for(int i = 1; i <= n; i++)
if(f[i] == k) ans++;
}
int main(){
fread(buf, sizeof(char), sizeof(buf), stdin);
k = readint();
n = readint();
m = readint();
for(int i = 1; i <= k; i++) w[i] = readint();
for(int u, v, i = 1; i <= m; i++){
u = readint();
v = readint();
add(u, v);
}
work();
printf("%d\n", ans);
return 0;
}