poj3071(概率dp)

最新推荐文章于 2019-09-09 19:15:59 发布

qkoqhh

最新推荐文章于 2019-09-09 19:15:59 发布

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分类专栏： DP

本文链接：https://blog.youkuaiyun.com/qkoqhh/article/details/79347613

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本文介绍了一种使用概率矩阵预测单淘汰赛制中哪个队伍最有可能获胜的方法。通过递归分解比赛过程，计算每个队伍赢得整个比赛的概率。

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这个就是纯概率问题了。。直接把对手的胜率乘上自己的胜率再乘自己打赢对方的概率就行。。。

题目设计成分治的样子，用分治写就很舒服了。。

/**
 *　　　　　　　　┏┓　　 　┏┓ 
 * 　　　　　　　┏┛┗━━━━━━━┛┗━━━┓
 * 　　　　　　　┃　　　　　　　┃ 　 
 * 　　　　　　　┃　　　━　　 　┃ 
 * 　　　　　　　┃　＞　　　＜　┃ 
 * 　　　　　　　┃　　　　　　　┃ 
 * 　　　　　　　┃...　⌒　... 　┃ 
 * 　　　　　　　┃　　　　　　　┃ 
 * 　　　　　　　┗━┓　　　┏━┛ 
 * 　　　　　　　　　┃　　　┃　Code is far away from bug with the animal protecting　　　　　　　　　　 
 * 　　　　　　　　　┃　　　┃   神兽保佑,代码无bug 
 * 　　　　　　　　　┃　　　┃　　　　　　　　　　　 
 * 　　　　　　　　　┃　　　┃  　　　　　　 
 * 　　　　　　　　　┃　　　┃ 
 * 　　　　　　　　　┃　　　┃　　　　　　　　　　　 
 * 　　　　　　　　　┃　　　┗━━━┓ 
 * 　　　　　　　　　┃　　　　　　　┣┓ 
 * 　　　　　　　　　┃　　　　　　　┏┛ 
 * 　　　　　　　　　┗┓┓┏━┳┓┏┛ 
 * 　　　　　　　　　　┃┫┫　┃┫┫ 
 * 　　　　　　　　　　┗┻┛　┗┻┛ 
 */  
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-8
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 205
#define nm 105
#define pi 3.1415926535897931
using namespace std;
const ll inf=1000000007;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}





int ans,n,m;
double c[NM],d[NM],a[NM][NM];

void dp(int x,int y){
	if(x==y){d[x]=1;return;}
	dp(x,mid);dp(mid+1,y);
	inc(i,x,y)c[i]=0;
	inc(i,x,mid)inc(j,mid+1,y)c[i]+=d[j]*a[i][j];
	inc(i,mid+1,y)inc(j,x,mid)c[i]+=d[j]*a[i][j];
	inc(i,x,y)d[i]*=c[i];
}

int main(){
	while((n=read())>0){
		m=succ(n);
		inc(i,1,m)inc(j,1,m)scanf("%lf",&a[i][j]);
		dp(1,m);ans=1;
		inc(i,2,m)if(d[i]>d[ans])ans=i;
		//inc(i,1,m)printf("%lf ",d[i]);
		printf("%d\n",ans);
	}
	return 0;
}

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6421		Accepted: 3279

Description

Consider a single-elimination football tournament involving 2ⁿ teams, denoted 1, 2, …, 2ⁿ. In each round of the tournament, all teams still in the tournament are placed in a list in order of increasing index. Then, the first team in the list plays the second team, the third team plays the fourth team, etc. The winners of these matches advance to the next round, and the losers are eliminated. After n rounds, only one team remains undefeated; this team is declared the winner.

Given a matrix P = [p_ij] such that p_ij is the probability that team i will beat team j in a match determine which team is most likely to win the tournament.

Input

The input test file will contain multiple test cases. Each test case will begin with a single line containing n (1 ≤ n ≤ 7). The next 2ⁿ lines each contain 2ⁿ values; here, the jth value on the ith line represents p_ij. The matrix P will satisfy the constraints that p_ij = 1.0 − p_ji for all i ≠ j, and p_ii = 0.0 for all i. The end-of-file is denoted by a single line containing the number −1. Note that each of the matrix entries in this problem is given as a floating-point value. To avoid precision problems, make sure that you use either the double data type instead of float.

Output

The output file should contain a single line for each test case indicating the number of the team most likely to win. To prevent floating-point precision issues, it is guaranteed that the difference in win probability for the top two teams will be at least 0.01.

Sample Input

2
0.0 0.1 0.2 0.3
0.9 0.0 0.4 0.5
0.8 0.6 0.0 0.6
0.7 0.5 0.4 0.0
-1

Sample Output

Hint

In the test case above, teams 1 and 2 and teams 3 and 4 play against each other in the first round; the winners of each match then play to determine the winner of the tournament. The probability that team 2 wins the tournament in this case is:

P(2 wins)