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Aizu 2303 Marathon Match 概率dp
题目链接：点击打开链接
题意：
一场马拉松
给定n个运动员，跑道上有m个休息站，马拉松跑道长L
下面n行每行3个参数
表示每个运动员崩溃概率:P 休息时间:E 跑步速度:V。
每个运动员随时会崩溃，崩溃后会坚持到下一个休息站，进入休息他的休息时间。全程匀速运动，可能多次崩溃。
问：
每个运动员成为唯一一个第一名的概率。
思路：
其实可以转换成每个运动员进入下一个休息站的概率是P
每次单独求一个运动员成为第一名的概率。
暴力枚举这个运动员进入休息站的次数。
然后枚举其他的运动员比他慢的概率，相乘得到这个运动员崩溃j次后依然是第一名的概率，相加即是他成为第一名的概率。
#include<bits/stdc++.h>
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if(c=getchar(),c==EOF) return 0;
while(c!='-'&&(c<'0'||c>'9')) c=getchar();
sgn=(c=='-')?-1:1;
ret=(c=='-')?0:(c-'0');
while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
ret*=sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x <0) {
putchar('-');
x = -x;
}
if(x>9) pt(x/10);
putchar(x%10+'0');
}
using namespace std;
typedef long long ll;
const int MAXN = 55;
double C[MAXN+1][MAXN+1];
void Initial()
{
memset(C, 0, sizeof C);
for(int i=0; i<=MAXN; ++i)
{
C[0][i] = 0.0;
C[i][0] = 1.0;
}
for(int i=1; i<=MAXN; ++i)
{
for(int j=1; j<=MAXN; ++j)
C[i][j] = C[i-1][j] + C[i-1][j-1];
}
}
const int N = 110;
int n, m;
int l, t[N], v[N];
double p[N];
double Pow(double x, int y){
double ans = 1.0;
while(y){
if(y&1)ans *= x;
x = x*x;
y >>= 1;
}
return ans;
}
double work(int x){
double ans = 0.0;
for(int i = 0; i <= m; i++)
{
double pp = C[m][i] * Pow(p[x], i) * Pow(1.0-p[x], m-i);
for(int j = 1; j <= n; j++)
{
if(j == x)continue;
double siz = 0;
for(int k = m; k >= 0; k--)
{
if(l*v[j] + i*t[x]*v[j]*v[x] < l*v[x] + k*t[j]*v[x]*v[j])
siz += C[m][k] * Pow(p[j], k) * Pow(1.0-p[j], m-k);
else break;
}
pp *= siz;
}
ans += pp;
}
return ans;
}
void input(){
for(int i = 1; i <= n; i++)
{
rd(p[i]);rd(t[i]);rd(v[i]);
p[i] /= 100.0;
}
}
int main(){
Initial();
while(cin>>n>>m>>l){
input();
for(int i = 1; i <= n; i++)
printf("%.10f\n", work(i));
}
return 0;
}
/*
30 0 12
0.5
0
13
*/Marathon Match
Time Limit : 5 sec, Memory Limit : 65536 KBMarathon 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≤N≤100), M (0≤M≤50) and L(1≤L≤100,000). 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≤Pi≤100), Ti (0≤Ti≤100) and Vi(0≤Vi≤100) 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
3 1 100 50 1 1 50 1 1 50 1 1
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
Aizu 2303 Marathon Match 概率dp