poj 3071 概率DP

http://poj.org/problem?id=3071

推方程不难，但是难在怎么算

dp[i][j]表示第i场时第j只队伍存活下来的概率 

方程：dp[i][j]=sigma(dp[i-1][j]*p[j][k]*dp[i-1][k])

j,k在同一场的条件：if(((k>>(i-1))^1)==(j>>(i-1)))即判断k的第i位前的数没有比过的是否与j的在同一棵子树上。（i从1取，j,k从0取）

题解参考 http://blog.csdn.net/pbj1203/article/details/6950450

//#pragma comment(linker, "/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
#include <iomanip>
#include <cmath>
#include <map>
#include <set>
#include <queue>
using namespace std;

#define ls(rt) rt*2
#define rs(rt) rt*2+1
#define ll long long
#define ull unsigned long long
#define rep(i,s,e) for(int i=s;i<e;i++)
#define repe(i,s,e) for(int i=s;i<=e;i++)
#define CL(a,b) memset(a,b,sizeof(a))
#define IN(s) freopen(s,"r",stdin)
#define OUT(s) freopen(s,"w",stdout)
const ll ll_INF = ((ull)(-1))>>1;
const double EPS = 1e-8;
const int INF = 100000000;
const int MAXN = 1000;

double p[MAXN][MAXN],dp[MAXN][MAXN];
int N,n;

int solve()
{
     CL(dp,0);
     rep(i,0,n)/////////
        dp[0][i]=1;
    repe(i,1,N)
        rep(j,0,n)
            rep(k,0,n)
                if(((k>>(i-1))^1) == (j>>(i-1)))
                    dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];
    int ans=-1;
    dp[0][0]=-1;
    rep(j,0,n)
        if(dp[N][j]>dp[0][0])
        {
             ans=j;
             dp[0][0]=dp[N][j];
        }

    return ans+1;

}

int main()
{
    while(~scanf("%d",&N) && N!=-1)
    {
        n=1<<N;
        rep(i,0,n)
            rep(j,0,n)
                scanf("%lf",&p[i][j]);

        printf("%d\n",solve());

    }
    return 0;
}