Cow Picnic
Time Limit: 2000 ms Memory Limit: 64 MB
Total Submission: 1 Accepted: 1
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.
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.
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
Hint
The cows can meet in pastures 3 or 4.
Source
USACO 2006 December Silver
從每個(gè)牛開始求一次單源最短路徑，假設(shè)起點(diǎn)是X，如果從X能到i (di[i]!=INF) ，cnt[i]++，用來(lái)統(tǒng)計(jì)能到達(dá) i 點(diǎn)的牛的數(shù)量。
結(jié)果就是滿足cnt[i]==K的數(shù)量，即i點(diǎn)所有的牛都可以到達(dá)。
用spfa求，spfa在這里不是求最段路徑，只要到了就行，不需要是最短的，因此會(huì)更快一點(diǎn)。
#include<iostream>
#include<time.h>
#include<vector>
#include<queue>
using namespace std;
const int MAX=1001,INF=0x0fffffff;
vector<int> mp[MAX];
int d[MAX], cnt[MAX];
int K,N,M;
int stay[101];
void spfa(int x)
{
for(int i=1; i<=N; i++)
d[i]=INF;
queue<int>q;
q.push(x);
d[x]=0;
while(q.size())
{
int u=q.front(); q.pop();
for(int i=0; i<mp[u].size(); i++)
{
if(d[mp[u][i]]==INF)
{
d[mp[u][i]]=d[u]+1;
q.push(mp[u][i]);
}
}
}
}
int main()
{
cin>>K>>N>>M;
for(int i=1; i<=K; i++)
cin>>stay[i];
for(int i=1,s,t; i<=M; i++)
{
cin>>s>>t;
mp[s].push_back(t);
}
for(int i=1; i<=K; i++)
{
spfa(stay[i]);
for(int i=1; i<=N; i++)
if(d[i]!=INF)cnt[i]++;
}
int ans=0;
for(int i=1; i<=N; i++)
if(cnt[i]==K)ans++;
cout<<ans<<endl;
system("pause");
return 0;
}