zoukankan      html  css  js  c++  java
  • CodeForces

    Discription

    n fish, numbered from 1 to n, live in a lake. Every day right one pair of fish meet, and the probability of each other pair meeting is the same. If two fish with indexes i and j meet, the first will eat up the second with the probability aij, and the second will eat up the first with the probability aji = 1 - aij. The described process goes on until there are at least two fish in the lake. For each fish find out the probability that it will survive to be the last in the lake.

    Input

    The first line contains integer n (1 ≤ n ≤ 18) — the amount of fish in the lake. Then there follow n lines with n real numbers each — matrix aaij (0 ≤ aij ≤ 1) — the probability that fish with index i eats up fish with index j. It's guaranteed that the main diagonal contains zeros only, and for other elements the following is true: aij = 1 - aji. All real numbers are given with not more than 6 characters after the decimal point.

    Output

    Output n space-separated real numbers accurate to not less than 6 decimal places. Number with index i should be equal to the probability that fish with index i will survive to be the last in the lake.

    Examples

    Input
    2
    0 0.5
    0.5 0
    Output
    0.500000 0.500000 
    Input
    5
    0 1 1 1 1
    0 0 0.5 0.5 0.5
    0 0.5 0 0.5 0.5
    0 0.5 0.5 0 0.5
    0 0.5 0.5 0.5 0
    Output
    1.000000 0.000000 0.000000 0.000000 0.000000 


    不难想到设f[S]为到达状态S下的概率,模拟鱼相遇的过程就可以做到 O(2^N * N^2) 的复杂度,足够通过本题。
    但是这还不是最优的方法,因为一个状态 S 的后继只有 O(N)种,并且在这个题中不同的到达后继的方式是很好合并的,根据每一对鱼之间相遇的概率的独立性,我们可以O(2^N * N)预处理出每个鱼i在集合S中被吃掉的概率 f[S][i],并通过这个直接从S -> S^(2^i) ,总的复杂度就是 O(2^N * N)。

    (假装我是CF上的rank1 23333)


    #include<bits/stdc++.h>
    #define ll long long
    using namespace std;
    #define D double
    const int maxn=400005;
    D a[23][23],f[maxn],BE[maxn][23];
    int ci[35],n,m,T,BT[maxn],dy[maxn];
    
    inline void prework(){
    	for(int i=1,now,lef;i<ci[n];i++){
    		now=i&-i,lef=i^now,now=dy[now];
    		for(int j=0;j<n;j++) BE[i][j]=BE[lef][j]+a[now][j];
    	}
    }
    
    inline void solve(){
    	f[ci[n]-1]=1;
    	for(int i=ci[n]-1;i;i--) if(BT[i]>1){
    		T=BT[i]*(BT[i]-1)>>1;
    		for(int j=0;j<n;j++) if(ci[j]&i) f[i^ci[j]]+=f[i]*BE[i][j]/(D)T;
    	}
    }
    
    int main(){
    	ci[0]=1; for(int i=1;i<=20;i++) ci[i]=ci[i-1]<<1;
    	
    	scanf("%d",&n);
    	for(int i=0;i<n;i++)
    	    for(int j=0;j<n;j++) scanf("%lf",&a[i][j]);
    	    
    	BT[0]=0; for(int i=1;i<ci[n];i++) BT[i]=BT[i^(i&-i)]+1;
    	for(int i=0;i<n;i++) dy[ci[i]]=i;
    	
    	prework(),solve();
    	
    	for(int i=0;i<n;i++) printf("%.10lf ",f[ci[i]]);
    	return 0;
    }
    

      

     
  • 相关阅读:
    CollectionView Header And Footer 的使用
    UICollectionView 的添加自定义高度的区头
    利用SDWebImage 清理缓存
    iOS tableView 中 UITableView中UITableViewStyleGrouped和UITableViewStylePlain的区别
    关于tableview的分割线隐藏问题
    改变输入框的placehould的颜色 和边框颜色
    将代码在有xib得情况下设置到最底层(即执行顺序不是先执行xib) 将图片毛玻璃状态
    通知监听键盘高度变化 自适应键盘高度
    iOS 定义枚举
    iOS 快速存储到本地
  • 原文地址:https://www.cnblogs.com/JYYHH/p/8983953.html
Copyright © 2011-2022 走看看