Explosion（bitset 优化的传递闭包 + 概率）

Explosion

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 259    Accepted Submission(s): 67

Problem Description

Everyone knows Matt enjoys playing games very much. Now, he is playing such a game. There are N rooms, each with one door. There are some keys(could be none) in each room corresponding to some doors among these N doors. Every key can open only one door. Matt has some bombs, each of which can destroy a door. He will uniformly choose a door that can not be opened with the keys in his hand to destroy when there are no doors that can be opened with keys in his hand. Now, he wants to ask you, what is the expected number of bombs he will use to open or destroy all the doors. Rooms are numbered from 1 to N.

 

 

Input

The first line of the input contains an integer T, denoting the number of testcases. Then T test cases follow.

In the first line of each test case, there is an integer N (N<=1000) indicating the number of rooms.

The following N lines corresponde to the rooms from 1 to N. Each line begins with an integer k (0<=k<=N) indicating the number of keys behind the door. Then k integers follow corresponding to the rooms these keys can open.

 

 

Output

For each test case, output one line "Case #x: y", where x is the case number (starting from 1), y is the answer which should be rounded to 5 decimal places.

 

 

Sample Input

2

3

1 2

1 3

1 1

3

0

0

0

 

 

Sample Output

 

Case #1: 1.00000

Case #2: 3.00000

 

       题意：

       给出 T 组 case，后给出 N 个房间，后有 N 行，第 i 行代表打开 i 号门后拥有的钥匙有什么，首先给出 K 代表有 K 条钥匙，后给出 K 条钥匙是什么。如果没有钥匙能打开这扇门则可以用  boom 炸开它，求能打开所有门所有 boom 数的数学期望。

 

        思路：

        统计 S 表示该扇门能被打开的房间总数有多少个，则 1 / S 则表示这扇门能被打开的概率，则 Sum { 1 / Si （1 <= i <= n ） } 即为所求。则题目转化为求传递闭包问题，用 bitset 优化可节省时间。

 

        AC：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <bitset>

using namespace std;

const int VMAX = 1005;

bitset<VMAX> Map[VMAX];
double res;
int n;

void floyd () {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (Map[j][i]) Map[j] |= Map[i];
        }
    }

    for (int i = 1; i <= n; ++i) {
        int ans = 0;
        for (int j = 1; j <= n; ++j) {
            if (Map[j][i]) ++ans;
        }

        res += 1.0 / (double) ans;
    }
}

int main() {

    int t;
    scanf("%d", &t);

    for (int tt = 1; tt <= t; ++tt) {

        scanf("%d", &n);

        res = 0;
        for (int i = 1; i <= n; ++i) {
            Map[i].reset();
            Map[i][i] = 1;
        }

        for (int i = 1; i <= n; ++i) {
            int ans;
            scanf("%d", &ans);
            while (ans--) {
                int j;
                scanf("%d", &j);
                Map[i][j] = 1;
            }
        }

        floyd();

        printf("Case #%d: %.5lf\n", tt, res);
    }

    return 0;
}