【BZOJ】1648: [Usaco2006 Dec]Cow Picnic 奶牛野餐(dfs)




#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1005, C=105;
int c, cow[C], n, ihead[N], cnt, m;
bool vis[C][N];
struct ED { int to, next; }e[10005];
void add(int u, int v) { e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; }
void dfs(bool *v, int x) {
	for(int i=ihead[x]; i; i=e[i].next) if(!v[e[i].to]) dfs(v, e[i].to);

int main() {
	read(c); read(n); read(m);
	for1(i, 1, c) read(cow[i]);
	while(m--) { int u=getint(), v=getint(); add(u, v); }
	for1(i, 1, c) dfs(vis[i], cow[i]);
	int ans=0;
	for1(i, 1, n) {
		bool v=1;
		for1(j, 1, c) v=v&&vis[j][i];
		if(v) ++ans;
	return 0;






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.



* 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.



* 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
1 2
1 4
2 3
3 4


^ ^
| |
| |

The pastures are laid out as shown above, with cows in pastures 2 and 3.

Sample Output





posted @ 2014-09-05 12:15  iwtwiioi  阅读(270)  评论(0编辑  收藏  举报