(简单) 树形dp+概率 POJ 3071 Football
2014-02-08 09:52
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Football
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 doubledata 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 Input2 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 Output2 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:
winning the tournament. Source Stanford Local 2006 |
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); }}
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