HDU 5036 Explosion (probability expectation +bitset)

problem Description

 is inch inch inch  is  from 1 to N.

Input

 Case  is an integer N (n<=-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

 Case " Case #x: y " where  is  Case  from 1  is 5 decimal places.

Sample Input

2 3 1 2 1 3 1 1 3 0 0 0

Sample Output

Case #11.00000 case #23.00000

Source ACM/ICPC Asia regional Beijing Online Test Instructions: a person to open or use a bomb smashed all the doors, each door there are some keys, a key to a door, a door with a key to open the corresponding door, tell each door what the key of the door, and asked how many bombs to use.   Ideas: considering the probability that each point needs to be opened with a bomb, the sum of the probabilities of all points is the solution. The probability that each point v needs to be opened with a bomb is 1/s, S is the number of U (u->v unicom), and then it becomes the transitive closure of this diagram, optimized with Bitset. Why is 1/s, because this S door inside must use a bomb to break open, and hit v probability is 1/s.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <algorithm>6#include <stdlib.h>7#include <bitset>8 using namespacestd;9 #defineN 1006Ten intN; OneBitset<n>Bs[n]; A intMain () - { -    intT; the    intAc=0; -scanf"%d",&t); -     while(t--){ -scanf"%d",&n); +        for(intI=0; i<n;i++){ - Bs[i].reset (); +bs[i][i]=true; A       } at        for(intI=0; i<n;i++){ -          intK; -scanf"%d",&k); -           for(intj=0; j<k;j++){ -             intx; -scanf"%d",&x); inx--; -bs[i][x]=true; to          } +       } -        for(intj=0; j<n;j++){ the           for(intI=0; i<n;i++){ *             if(Bs[i][j]) { $Bs[i] |=Bs[j];Panax Notoginseng             } -          } the       } +       Doubleans=0; A        for(intj=0; j<n;j++){ the          intCnt=0; +           for(intI=0; i<n;i++){ -             if(Bs[i][j]) { $cnt++; $             } -          } -ans+=1.0/CNT; the       } -printf"Case #%d:%.5lf\n",++Ac,ans);Wuyi    } the    return 0; -}

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HDU 5036 Explosion (probability expectation +bitset)