(简单) 树形dp+概率 POJ 3071 Football

Football

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 2433Accepted: 1229

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

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.

Source

Stanford Local 2006

题意：就是想天下第一武道会那样选出优胜者.......给出2^n个选手 按1~2^n 排列， 然后1和2比赛，3和4比赛......选出优胜者，然后在这些优胜者同样继续这样较量，最后会有一个人胜出。该题给出i和j比赛时，i打败j的概率，求哪一个选手最有机会获胜。

思路：我们先建立一个完全二叉树，叶子节点从左到右分别是1,2,3,4.......2^n，其他节点编号任意。然后从这颗树的底部往上更新，我们用dp[i][j] 表示在i节点胜出者是j的概率 用lson 表示i的左儿子，rson 表示i的右儿子  那么dp[i][j] = ∑(dp[lson][j]*dp[rson][k]+dp[rson][k]*dp[rson][j])*p[j][k] , (1<=k<=2^n && k!=j) 

代码：

#include<iostream>

#include<stdio.h>

#include<cstdio>

#include<algorithm>

#include<math.h>

#include<string.h>

using namespace std;

const int maxn = 1000+10;

const int MOD = 1000000007;

const int inf = 0xfffffff;

typedef  long long LL;

double p[maxn][maxn];

double dp[maxn][maxn];

int n;

int sz;

int cnt;

struct Node

{

int lson;

int rson;

}node[maxn];

void build(int dep,int rt)

{

if (dep==n-1)

{

node[sz].lson = cnt++;

node[sz].rson = cnt++;

return;

}

node[rt].lson = ++sz;

build(dep+1,sz);

node[rt].rson = ++sz;

build(dep+1,sz);

}

void tp(int rt)

{

int ls = node[rt].lson;

int rs = node[rt].rson;

if (ls==-1)

{

dp[rt][rt] = 1;

return;

}

tp(ls) , tp(rs);

for (int i = 1 ; i <= (1<<n) ; ++i)

{

for (int j = 1 ; j <= (1<<n) ; ++j) 

dp[rt][i] += (dp[rs][i]*dp[ls][j]+dp[ls][i]*dp[rs][j])*p[i][j];

}

}

int main()

{

while (scanf("%d",&n))

{

if (n==-1) return 0;

memset(node,-1,sizeof(node));

for (int i = 1 ; i <= (1<<n) ; ++i)

for (int j = 1 ; j <= (1<<n) ; ++j)

scanf("%lf",&p[i][j]);

sz = 1<<n|1;

cnt = 1;

build(0,sz);

memset(dp,0,sizeof(dp));

tp(1<<n|1);

int ret = 1;

for (int i = 2 ; i <= (1<<n) ; ++i)

if (dp[1<<n|1][i] > dp[1<<n|1][ret])

ret = i;

printf("%d\n",ret);

}

}