标签:
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. 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 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:
The next most likely team to win is team 3, with a 0.372 probability of winning the tournament. Source |
设dp[i][left][right]表示i这支球队在第left到第right球队里面得到冠军的概率,所以最后需要的是dp[i][1][16]。一个球队要在当前的阶段得到冠军必须在之前得到冠军
所以mid=(left+right)/2;
i和j在[left,right]里面争夺冠军(例如i和j在1到16里面争夺冠军,也就是争夺总冠军),那么i和j必定是来自于两个分区,left<=i<=mid和mid+1<=j<=right
即状态转移方程:dp[j][l][r]+=beat[j][k]*dp[j][l][mid]*dp[k][mid+1][r];
dp[k][l][r]+=beat[k][j]*dp[j][l][mid]*dp[k][mid+1][r];
//17356K 454MS #include <stdio.h> #include <math.h> #include <string.h> using namespace std; double beat[129][129],dp[129][129][129]; int main() { int n; while(scanf("%d",&n)&&n!=-1) { n=(int)pow(2,n); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) scanf("%lf",&beat[i][j]); int r,mid; memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++)dp[i][i][i]=1; for(int i=2;i<=n;i<<=1)//第几次淘汰赛 for(int l=1;l<=n;l+=i)//每次淘汰赛的左区间 { r=l+i-1;//每次淘汰赛的右区间 mid=(l+r)>>1; for(int j=l;j<=mid;j++) for(int k=mid+1;k<=r;k++) { dp[j][l][r]+=beat[j][k]*dp[j][l][mid]*dp[k][mid+1][r]; dp[k][l][r]+=beat[k][j]*dp[j][l][mid]*dp[k][mid+1][r]; } } double maxx=-1; int pos; for(int i=1;i<=n;i++) { if(dp[i][1][n]>maxx) { maxx=dp[i][1][n]; pos=i; } } printf("%d\n",pos); } return 0; }
//528K 79MS #include<stdio.h> #include<string.h> double p[129][129],dp[17][129]; int main() { int n; while(scanf("%d",&n)&&n!=-1) { for(int i=0;i<(1<<n);i++) for(int j=0;j<(1<<n);j++) scanf("%lf",&p[i][j]); memset(dp,0,sizeof(dp)); for(int i=0;i<(1<<n);i++)dp[0][i]=1; for(int i=1;i<=n;i++) for(int j=0;j<(1<<n);j++) for(int k=0;k<(1<<n);k++) if(((j>>(i-1))^1)==(k>>(i-1))) dp[i][j]+=dp[i-1][j]*dp[i-1][k]*p[j][k]; double maxx=-1; int pos; for(int i=0;i<(1<<n);i++) { if(dp[n][i]>maxx) { maxx=dp[n][i]; pos=i+1; } } printf("%d\n",pos); } return 0; }
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原文地址:http://blog.csdn.net/crescent__moon/article/details/43272349