POJ - 3071 Football（概率DP）

题目链接
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 = [pij] such that pij 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 pij. The matrix P will satisfy the constraints that pij = 1.0 − pji for all i ≠ j, and pii = 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
2
0.0 0.1 0.2 0.3
0.9 0.0 0.4 0.5
0.8 0.6 0.0 0.6
0.7 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.
题意：
有2^n支队，现在要进行n次比赛，并且按次序进行比赛并淘汰，胜利的队继续按次序比赛并淘汰，比如1，2，3，4进行比赛，第一轮1和2比，3和4比，假如1和3胜利了，那么第二轮1和3继续比，2，4淘汰。最后问最有可能胜利的队伍是哪一支。，p【i】【j】代表的是i队伍打败j队伍的概率。
思路：dp【i】【j】代表第i轮j能取胜的概率，如果j要取胜的话dp【i-1】【j】肯定也必须要取胜，同时和j对决的对手k也得能活着和它比赛才性，于是转移就是dp【i】【j】+=dp【i-1】【j】*（p【j】【k】 *dp【i-1】【k】），累加的话其实就是全概率公式，剩下的就是dp的递推计算了。

#include
#include
using namespace std;
double dp[305][305],p[305][305];
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==-1) break;
		int num=(1<<n),ans=0;
		memset(dp,0,sizeof(dp));
		for(int i=0;i<num;++i) dp[0][i]=1;
		for(int i=0;i<num;++i)
		for(int j=0;j<num;++j)
		scanf("%lf",&p[i][j]);
		for(int i=1;i<=n;++i)
		for(int j=0;j<num;++j)
		for(int k=0;k<num;++k)
		if((j>>(i-1))==((k>>(i-1))^1))//这里注意一下，这里是判断j和k是那支队伍的并且是不是对手
		dp[i][j]+=dp[i-1][j]*(p[j][k]*dp[i-1][k]);
		for(int i=1;i<num;++i)
		if(dp[n][i]>dp[n][ans]) ans=i;
		printf("%d\n",ans+1);
	}
 }