HDU 5036 Explosion (bitset + DP)

题意：

给你一张图，每个图都可以到达另一个节点，当然你也可以选择炸开者结点来进入这个图，问你炸开节点的期望次数是多少。

分析：

因为每个点是独立的，我们考虑每一个点的单独的期望，到这个点要炸的次数其实就是到这个点的节点数的倒数，所以我们需要把可以到达每个点的点数处理出来，直接求的话，用floyd来求闭包，复杂度不够，所以用bitset来优化。

代码：

//
//  Created by  CQU_CST_WuErli
//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.
//
//#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <string>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <set>
#include <algorithm>
#include <sstream>
#include <bitset>
#define CLR(x) memset(x,0,sizeof(x))
#define OFF(x) memset(x,-1,sizeof(x))
#define MEM(x,a) memset((x),(a),sizeof(x))
#define BUG cout << "I am here" << endl
#define lookln(x) cout << #x << "=" << x << endl
#define SI(a) scanf("%d", &a)
#define SII(a,b) scanf("%d%d", &a, &b)
#define SIII(a,b,c) scanf("%d%d%d", &a, &b, &c)
const int INF_INT=0x3f3f3f3f;
const long long INF_LL=0x7f7f7f7f;
const int MOD=1e9+7;
const double eps=1e-10;
const double pi=acos(-1);
typedef long long  ll;
using namespace std;

const int N = 1010;

int n;
bitset<N> bit
;

int main(int argc, char const *argv[]) {
#ifdef LOCAL
freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);
// freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);
#endif
for (int T_T, kase = SI(T_T); kase <= T_T; kase++) {
SI(n);
for (int i = 1; i <= n; i++)
bit[i].reset();
for (int i = 1; i <= n; i++) {
int x; SI(x);
bit[i].set(i);
while (x--) {
int pos; SI(pos);
bit[i].set(pos);
}
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) if (bit[j][i]) {
bit[j] |= bit[i];
}
}
double ans = 0.0;
for (int i = 1; i <= n; i++) {
double tmp = 0.0;
for (int j = 1; j <= n; j++) if (bit[j][i])
tmp += 1.0;
ans += 1.0 / tmp;
}
printf("Case #%d: %.5f\n", kase, ans);
}
return 0;
}