POJ 3071 Football—概率DP-conquer sea博客

                                                                                    POJ 3071 Football

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. Aftern 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 teami 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 containingn (1 ≤ n ≤ 7). The next 2ⁿ lines each contain 2ⁿ values; here, thejth value on the ith line represents p_ij. The matrixP will satisfy the constraints that p_ij = 1.0 − p_ji for all i ≠ j, and p_ii = 0.0 for alli. 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

这道题其实比较容易思考，把每一轮某人获胜的概率记录就行了。

dp[i][j]表示第J人i轮出线的概率

麻烦的是判断这轮有哪几个可能和自己交手（k循环），把和这几个人打的胜率*这几个人本轮出线胜率相乘再累加就行了。

#include <cstring>#include <cstdio>using namespace std;double dp[8][130],p[130][130];int n,m;int main(){    while(~scanf("%d",&n)&&n!=-1)    {        m=1<<n;        for(int i=1; i<=m; i++)            for(int j=1; j<=m; j++)                scanf("%lf",&p[i][j]);                memset(dp,0,sizeof(dp));        for(int i=1;i<=m;i++)           dp[0][i]=1;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)                for(int k=1;k<=m;k++)                    if(((k-1)>>i)==(j-1)>>i&&!((k-1)>>(i-1)==(j-1)>>(i-1)))                    dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k];        int ans=1;        for(int i=2;i<=m;i++)        if(dp[n][i]>dp[n][ans])            ans=i;            printf("%d\n",ans);    }    return 0;}