Football

Subject:
https://ac.nowcoder.com/acm/problem/107668

Gives you $2^n$ teams, adjacent teams play in each round. Tell you$i$ team beatjjThe probability of team$j$ is$P_{ij}$, Ask which team has the greatest chance of winning!

Idea:
Let $f(i,j)$ means team$i$ can advance tojjProbability of round$j$ , initially$0$轮,$f(i,0)=1$ , and then for all possiblejjround$j$ and team$i$ match team$k$， 有
$$f(i,j)=\sum _{k}f(i,j-1)\ast P_{ik}\ast f(k,j-1)\; The$$
explanation is$i$ team$In$ round$j,$ advance to$j-1$ round and beat$j-$All possible opponents in$1$ round.
In addition, we can determine the number of team i in round j, for example
,$0$ rounds$1,2,3,4,5,6,7,8$
No.$1$ round$1,3,5,7$
of$2$ rounds$3,5$
,$3$ rounds$5$
Observe$5$ , will find the team$i\; is$ injjThe position of wheel$j$ is$\lceil \frac{i}{2^j}\rceil$ , equivalent to$j$ round with$2^j$ is a group, each group wins one person, then$i$ is the first$\lceil \frac{i}{2^j}\rceil Group$ .

• 如果$\lceil \frac{i}{2^j}\rceil$ is odd, then sum$\lceil \frac{i}{2^j}\rceil +1$ group play.
• 如果$\lceil \frac{i}{2^j}\rceil$ is even, then sum$\lceil \frac{i}{2^j}\rceil -1$ group play.

#include<stdio.h>
#include<cstring>
using namespace std;
const int N=8;
const int eps=1e-6;
double a[1<<N][1<<N],f[1<<N][N],maxi;
int n,locate;
double dfs(int x,int y)
{
    
    
    if(f[x][y])
        return f[x][y];
    f[x][y]=0;
    int tu=x%(1<<(y-1))?x/(1<<(y-1))+1:x/(1<<(y-1));//x为第几个
    tu=tu%2?tu+1:tu-1;//x和哪组打
    for(int i=1; i<=1<<n; i++)//枚举可能的对手
    {
    
    
        int tu1=i%(1<<(y-1))?i/(1<<(y-1))+1:i/(1<<(y-1));
        if(tu1==tu)
            f[x][y]+=a[x][i]*dfs(i,y-1);
    }
    f[x][y]*=dfs(x,y-1);
    return f[x][y];
}
int main()
{
    
    

    while(scanf("%d",&n)==1&&n!=-1)
    {
    
    
        for(int i=1; i<=1<<n; i++)
            for(int j=1; j<=1<<n; j++)
                scanf("%lf",&a[i][j]);
        for(int i=1; i<=1<<n; i++)
            f[i][0]=1.0;
        for(int i=1; i<=1<<n; i++)
        {
    
    
            double res=dfs(i,n);
            if(res-maxi>eps)
                maxi=res,locate=i;
        }
        printf("%d\n",locate);
    }
    return 0;
}