Marathon Match

最新推荐文章于 2025-06-10 09:04:25 发布

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最新推荐文章于 2025-06-10 09:04:25 发布

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分类专栏：概率DP 文章标签： ACM

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本文详细介绍了如何计算在设定的马拉松比赛中每个参赛者赢得第一的可能性，包括参赛人数、休息地点数量、选手跑步速度、休息时间等参数，并通过代码实现了这一算法。文章深入探讨了概率论在实际比赛场景中的应用。

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Marathon Match

N people run a marathon. There are M resting places on the way. For each resting place, the i-th runner takes a break with probability Pi percent. When the i-th runner takes a break, he gets rest for Ti time.

The i-th runner runs at constant speed Vi, and the distance of the marathon is L.

You are requested to compute the probability for each runner to win the first place. If a runner arrives at the goal with another person at the same time, they are not considered to win the first place.

Input

A dataset is given in the following format:

N M L
P1 T1 V1
P2 T2 V2
…
PN TN VN

The first line of a dataset contains three integers N (1¥leqN¥leq100), M (0¥leqM¥leq50) and L (1¥leqL¥leq100,000). N is the number of runners. M is the number of resting places. L is the distance of the marathon.

Each of the following N lines contains three integers Pi (0¥leqPi¥leq100), Ti (0¥leqTi¥leq100) and Vi (0¥leqVi¥leq100) describing the i-th runner. Pi is the probability to take a break. Ti is the time of resting. Vi is the speed.

Output

For each runner, you should answer the probability of winning. The i-th line in the output should be the probability that the i-th runner wins the marathon. Each number in the output should not contain an error greater than 10−5.

Sample Input 1

2 2 50
30 50 1
30 50 2

Output for the Sample Input 1

0.28770000
0.71230000

Sample Input 2

2 1 100
100 100 10
0 100 1

Output for the Sample Input 2

0.00000000
1.00000000

Sample Input 3

Output for the Sample Input 3

0.12500000
0.12500000
0.12500000

Sample Input 4

2 2 50
30 0 1
30 50 2

Output for the Sample Input 4

0.51000000
0.49000000

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define N 105
#define M 55
double C[M][M];
int n, m;
double L, P[N], T[N], V[N];

void pre_do()
{
    for(int i = 0; i <= 50; i++)
        for(int j = 0; j <= 50; j++)
            C[i][j] = 0;
    for(int i = 0; i <= 50; i++)
    {
        C[i][0] = 1;
        for(int j = 1; j <= i; j++)
            C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
    }
}

double mul_pow(double a, int k)
{
    double res = 1;
    while(k)
    {
        if(k & 1) res = res * a;
        a = a * a;
        k >>= 1;
    }
    return res;
}

double solve(int i, int k1, int j)
{
    if(V[j] == 0) return 1;
    double res = 0;
    for(int k2 = m; k2 >= 0; k2--)
    {
        if(L * V[j] + V[i] * V[j] * k1 * T[i] >= L * V[i] + V[i] * V[j] * k2 * T[j]) break;
        res += C[m][k2] * mul_pow(P[j], k2) * mul_pow(1 - P[j], m - k2);
    }
    return res;
}

int main()
{
    pre_do();
    while(~scanf("%d%d%lf", &n, &m, &L))
    {
        for(int i = 1; i <= n; i++)
        {
            scanf("%lf%lf%lf", &P[i], &T[i], &V[i]);
            P[i] /= 100;
        }
        double ans;
        for(int i = 1; i <= n; i++)
        {
            ans = 0;
            if(V[i] != 0)
            for(int k1 = 0; k1 <= m; k1++)
            {
                double tmp = C[m][k1] * mul_pow(P[i], k1) * mul_pow(1 - P[i], m - k1);
                for(int j = 1; j <= n; j++)
                if(j != i)
                {
                    tmp *= solve(i, k1, j);
                    if(tmp == 0) break;
                }
                ans += tmp;
            }
            printf("%.10f\n", ans);
        }
    }
    return 0;
}
/*
2 2 50
30 50 1
30 50 2

2 1 100
100 100 10
0 100 1

3 1 100
50 1 1
50 1 1
50 1 1

2 2 50
30 0 1
30 50 2
*/