poj3071（概率DP）

地址：http://poj.org/problem?id=3071

Football

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2559 Accepted: 1295

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

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

题意：求哪只足球队胜率最高。

思路：这是第二道概率DP，还是不清楚怎么求动态转移方程，或者说从哪里入手。个人感觉这道题做法是将所有情况穷举出来，结果真是这样，但是用转移方程推各种情况要比一般方法简单。转移方程：dp[i][j]=dp[i-1][j]*(p[j][h1]*dp[i-1][h1]+p[j][h2]*dp[i-1][h2]+......+p[j][hn]*dp[i-1][hn])，dp[i][j]是在第i轮j球队存货的概率，p[j][h]是j球队战胜h球队的概率。

代码：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;double p[200][200],dp[10][200];void getans(int n){    int m=1<<n;    for(int i=0;i<n;i++)    {        int k=1<<i,l,r;        for(int j=0;j<m;j++)        {            if((j/k)%2==1)                l=(j/k-1)*k,r=j/k*k;            else l=(j/k+1)*k,r=(j/k+2)*k;            for(int h=l;h<r;h++)            {                if(i==0) dp[i][j]=p[j][h];                else dp[i][j]+=dp[i-1][j]*p[j][h]*dp[i-1][h];            }        }    }}int main(){    int n;    while(scanf("%d",&n)>0,n!=-1)    {        int m=1<<n;        for(int i=0;i<m;i++)        {            for(int j=0;j<m;j++)                scanf("%lf",&p[i][j]);            for(int j=0;j<n;j++)                dp[j][i]=0;        }        getans(n);        int ans=0;double ansp=dp[n-1][0];        for(int i=1;i<m;i++)            if(dp[n-1][i]>ansp)            {                ansp=dp[n-1][i];                ans=i;            }        printf("%d\n",ans+1);    }    return 0;}