poj3071_Football 概率DP、二进制的应用

Football

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4987		Accepted: 2540

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 nrounds, 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

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

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:

扫描二维码关注公众号，回复： 5520725 查看本文章

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。

分析：

分析题目，先创建状态，dp[i][j]，表示在第i轮比赛中第j支队获胜，那么dp[0][j] = 1。然后考虑状态间的转移，很明显，如果j队要在本轮中获胜，前提是在上一轮中必须要先获胜才有资格，即dp[i-1][j]，并且在本轮中要击败所有可能的对手，那同时也要求对手也要在上一轮中获胜，才有资格进入本轮，即p[j][k]*dp[i-1][k]，其中p[j][k]代表j击败k的概率，所以状态转移方程为：dp[i][j] = dp[i-1][j] * (dp[i-1][k]*p[j][k])，其中k为本轮与j比赛的队伍。可是，不知道大家发现没有，这里有一个问题，那就是题目中说的淘汰制度，很明显必须要保证本轮比赛的双方曾经是没有比赛过的，因为如果两队曾经遇到过，必然会淘汰一队，那么现在又怎么可能再次比赛呢？所以在进行状态转移之前必须加一个条件，保证j和k是第一次进行比赛，那当我们将比赛流程用二进制表示时，会发现规律，当j>>(i-1) 等于 (k>>(i-1))^1时，j和k在第i轮比赛中第一次相遇。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
double p[150][150];
double dp[10][150];
int main(){
	int n;
	while(scanf("%d",&n)){
		if(n== -1)break;
		int num = 1 << n;
		for(int i = 0; i < num; i++)
			for(int j = 0; j < num; j++){
				scanf("%lf",&p[i][j]);
			}
        memset(dp,0,sizeof(dp));
        for(int i = 0; i < 140; i++)
            dp[0][i] = 1;
        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) ){
                        dp[i][j] += dp[i-1][j]*dp[i-1][k]*p[j][k];
                    }
                }
        double maxx = 0;
        int maxi;
        for(int i = 0; i < num; i++)
            if(dp[n][i] > maxx){
                maxx = dp[n][i];
                maxi = i+1;
            }
        printf("%d\n",maxi);
	}
}

poj3071_Football 概率DP、二进制的应用

猜你喜欢