1023: Pong’s Birds
时间限制: 1 Sec
内存限制: 128 MB
提交: 137
解决: 33
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题目描述
In order to train his birds, Pong is holding a competition for them. (He have birds, does he?) He have 2n
birds,
so he want to hold elimination series. Literally, if n = 2 and he has 4 birds identified as 1,2,3,4, he would first hold
competition for 1 and 2, 3 and 4, the one fails would be eliminated and then he holds competition for the winners.
According to his long observation, for each pair of birds he knows the probability to win for each bird, and he
want to know the probability of each bird to be the final winner.
输入
For the first row there is an integer T(T ≤ 100), which is the number of test cases.
For each case , first row is an integer n(1 ≤ n ≤ 7), for the next 2n
rows , each row has 2n
real numbers as the
probability to win for the i-th when competed to j-th bird. It is promised that for each i, j p[i][j] + p[j][i] = 1 and
p[i][i] = 0;
输出
For each case you should output a line with 2n
real numbers, which is the probability of i-th bird be the final
winner. Rounded to 3 decimal places.
Make sure to output a space character after each number, to prevent the Presentation Error.
样例输入
1
1
0 0.5
0.5 0
样例输出
0.500 0.500
分析：经典的dp题目；思路见代码；
代码：
#include<stdio.h>
#include<math.h>
double map[200][200];
double dp[200][200];
int main()
{
int T,n;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int maxn=pow(2,n);
for(int i=1;i<=maxn;i++)
{
for(int j=1;j<=maxn;j++)
scanf("%lf",&map[i][j]);
}
for(int i=1;i<=maxn;i++)
dp[0][i]=1.0;
int cnt;
for(int i=1;i<=n;i++) //一共有n轮比赛
{
for(int j=1;j<=maxn;j++) //求出每个人在第i轮比赛的取得胜利的概率，最后状态转移到总冠军
{
dp[i][j]=0.0;
int tem=j;
if(i>1)
tem=ceil((double)(j)/(double)pow(2,i-1));
cnt=0;
if(tem%2==1)
{
tem++; //tem和k计算出pow(2,i-1)个对手所在区间的起始位置
for(int k=tem*pow(2,i-1);cnt<pow(2,i-1);cnt++,k--) //pow(2,i-1)控制本轮可能遇到的对手的个数
dp[i][j] += dp[i-1][j]*map[j][k]*dp[i-1][k]; //晋级的概率*获胜的概率*对手晋级的概率
}
else
{
tem--;
for(int k=tem*pow(2,i-1);cnt<pow(2,i-1);cnt++,k--)
dp[i][j] += dp[i-1][j]*map[j][k]*dp[i-1][k];
}
}
}
printf("%.3f",dp[n][1]);
for(int id=2;id<=maxn;id++)
printf(" %.3f",dp[n][id]);
putchar('\n');
}
return 0;
}
