【传递闭包】 HDOJ 5036 Explosion

只要求每个门被开掉的数学期望和就行了。。。用floyd传递闭包。。。用bitset优化一下。。。

#include <iostream>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <bitset>  
#include <cstdio>  
#include <algorithm>  
#include <cstring>  
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 1005
#define maxm 300005
#define eps 1e-10
#define mod 10000007
#define INF 1e17
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
//head

bitset<maxn> p[maxn];
int n, m;

void read(void)
{
	scanf("%d", &n);
	for(int i = 1; i <= n; i++) {
		p[i].reset();
		p[i][i] = true;
	}
	for(int i = 1; i <= n; i++) {
		scanf("%d", &m);
		while(m--) {
			int j;
			scanf("%d", &j);
			p[i][j] = true;
		}
	}
}
void work(int _)
{
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			if(p[j][i]) p[j] |= p[i];
	double ans = 0;
	for(int i = 1; i <= n; i++) {
		int cnt = 0;
		for(int j = 1; j <= n; j++)
			if(p[j][i]) cnt++;
		ans += 1.0 / cnt;
	}
	printf("Case #%d: %.5f\n", _, ans);
}
int main(void)
{
	int _, __;
	while(scanf("%d", &_)!=EOF) {
		__ = 0;
		while(_--) {
			read();
			work(++__);
		}
	}
	return 0;
}