poj3256 Cow Picnic

Cow Picnic

Time Limit:2000MS	Memory Limit:65536K
Total Submissions:5811	Accepted:2388

Description

The cows are having a picnic! Each of Farmer John'sK(1 ≤K≤ 100) cows is grazing in one ofN(1 ≤N≤ 1,000) pastures, conveniently numbered 1...N. The pastures are connected byM(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.

Input

Line 1: Three space-separated integers, respectively:K,N, andM
Lines 2..K+1: Linei+1 contains a single integer (1..N) which is the number of the pasture in which cowiis grazing.
LinesK+2..M+K+1: Each line contains two space-separated integers, respectivelyAandB(both 1..NandA!=B), representing a one-way path from pastureAto pastureB.

Output

Line 1: The single integer that is the number of pastures that are reachable by all cows via the one-way paths.

Sample Input

2 4 4
2
3
1 2
1 4
2 3
3 4

Sample Output

2

Hint

The cows can meet in pastures 3 or 4.

Source

USACO 2006 December Silver

从每头牛开始的地方dfs一下。。。经过的点x全部cnt[x]++。。。然后看一下那些点的cnt值等于k。。。

用到了一个小trick。。。vis[i][j]表示点i是否第j头牛曾经经过。。。（空间换时间。。。）

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <cmath>
#include <cctype>
#include <string>
#include <cfloat>
#include <stack>
#include <cassert>

using namespace std;

int k, n, m, val[1010], cow[110], head[10010], rest[10010], to[10010], tot, vis[10010][110], ans;

void add(int u, int v) {
    tot ++;
    to[tot] = v;
    rest[tot] = head[u];
    head[u] = tot;
}

void dfs(int u, int ext) {
    vis[u][ext] = 1;
    val[u] ++;
    for(int i = head[u] ; i ; i = rest[i]) {
        int v = to[i];
        if(vis[v][ext]) continue;
        dfs(to[i], ext);
    }
}

int main() {
    scanf("%d%d%d", &k, &n, &m);
    for(int i = 1 ; i <= k ; i ++) {
        scanf("%d", &cow[i]);
    }
    for(int i = 1, a, b ; i <= m ; i ++) {
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    for(int i = 1 ; i <= k ; i ++) {
        dfs(cow[k], i);
    }
    for(int i = 1 ; i <= n ; i ++) {
        if(val[i] == k) {
            ans ++;
        }
    }
    printf("%d\n", ans);
}