POj 3071(概率dp)

题意：就是足球比赛，每个队伍对于不同队伍的胜利的概率是不一样的让你求出一个获胜的最大概率。

dp[i][j]表示：第i轮，第j队伍获胜的概率。

每支队伍这轮赢的前提是他上一轮获胜并且这一轮的对手上一轮也要获胜。

Football

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2803 Accepted: 1429

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 thedouble 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

20.0 0.1 0.2 0.30.9 0.0 0.4 0.50.8 0.6 0.0 0.60.7 0.5 0.4 0.0-1

Sample Output

2

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) = P(2 beats 1)P(3 beats 4)P(2 beats 3) + P(2 beats 1)P(4 beats 3)P(2 beats 4)
= p₂₁p₃₄p₂₃ + p₂₁p₄₃p₂₄
= 0.9 · 0.6 · 0.4 + 0.9 · 0.4 · 0.5 = 0.396.

The next most likely team to win is team 3, with a 0.372 probability of winning the tournament.

#include <algorithm>#include <iostream>#include <stdlib.h>#include <string.h>#include <iomanip>#include <stdio.h>#include <string>#include <queue>#include <cmath>#include <stack>#include <map>#include <set>#define eps 1e-7//#define M 1000100//#define LL __int64#define LL long long#define INF 0x3f3f3f3f#define PI 3.1415926535898const int maxn = 512;using namespace std;double p[maxn][maxn];double dp[maxn][maxn];int main(){    int n;    while(cin >>n)    {        if(n == -1)            break;        for(int i = 0; i < (1<<n); i++)            for(int j = 0; j < (1<<n); j++)                scanf("%lf",&p[i][j]);        memset(dp, 0, sizeof(dp));        for(int i = 0; i <= (1<<n); i++)            dp[0][i] = 1;        for(int i = 1; i <= n; i++)        {            for(int j = 0; j < (1<<n); j++)            {                int t=j/(1<<(i-1));                t^=1;                dp[i][j]=0;                for(int k=t*(1<<(i-1)); k < t*(1<<(i-1))+(1<<(i-1)); k++)                    dp[i][j] += (dp[i-1][j]*dp[i-1][k]*p[j][k]);            }        }        double Max = 0.0;        int x;        for(int i = 0; i < (1<<n); i++)        {            if(Max < dp[n][i])            {                x = i;                Max = dp[n][i];            }        }        cout<<x+1<<endl;    }    return 0;}