[hdu 4405] Aeroplane chess [概率DP & 期望]
2016-07-22 12:16
411 查看
[align=center]New~ 欢迎参加2016多校联合训练的同学们~
[/align]
Total Submission(s): 3221 Accepted Submission(s): 2048
[align=left]Problem Description[/align]
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are
1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is
no two or more flight lines start from the same grid.
Please help Hzz calculate the expected dice throwing times to finish the game.
[align=left]Input[/align]
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
[align=left]Output[/align]
For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.
[align=left]Sample Input[/align]
2 0
8 3
2 4
4 5
7 8
0 0
[align=left]Sample Output[/align]
1.1667
2.3441
[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Jinhua Online
[align=left]Recommend[/align]
zhoujiaqi2010 | We have carefully selected several similar problems for you: 5746 5745 5744 5743 5742
题目:http://acm.hdu.edu.cn/showproblem.php?pid=4405
题意:飞行棋,从0到n,置骰子,置到几就往前走几步,前进中会有捷径,比如2和5连到一起了,那你走到2时可以直接跳到
5,如果5和8连到一起了,那你还可以继续跳到8,最后问跳到n时平均置几次骰子。也就是求期望。
全期望公式:http://zh.wikipedia.org/wiki/%E5%85%A8%E6%9C%9F%E6%9C%9B%E5%85%AC%E5%BC%8F
全概率公式:http://zh.wikipedia.org/wiki/%E5%85%A8%E6%A6%82%E7%8E%87%E5%85%AC%E5%BC%8F
概率期望学习:http://kicd.blog.163.com/blog/static/126961911200910168335852/
ACcode:
[/align]
Aeroplane chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3221 Accepted Submission(s): 2048
[align=left]Problem Description[/align]
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are
1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is
no two or more flight lines start from the same grid.
Please help Hzz calculate the expected dice throwing times to finish the game.
[align=left]Input[/align]
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
[align=left]Output[/align]
For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.
[align=left]Sample Input[/align]
2 0
8 3
2 4
4 5
7 8
0 0
[align=left]Sample Output[/align]
1.1667
2.3441
[align=left]Source[/align]
2012 ACM/ICPC Asia Regional Jinhua Online
[align=left]Recommend[/align]
zhoujiaqi2010 | We have carefully selected several similar problems for you: 5746 5745 5744 5743 5742
题目:http://acm.hdu.edu.cn/showproblem.php?pid=4405
题意:飞行棋,从0到n,置骰子,置到几就往前走几步,前进中会有捷径,比如2和5连到一起了,那你走到2时可以直接跳到
5,如果5和8连到一起了,那你还可以继续跳到8,最后问跳到n时平均置几次骰子。也就是求期望。
全期望公式:http://zh.wikipedia.org/wiki/%E5%85%A8%E6%9C%9F%E6%9C%9B%E5%85%AC%E5%BC%8F
全概率公式:http://zh.wikipedia.org/wiki/%E5%85%A8%E6%A6%82%E7%8E%87%E5%85%AC%E5%BC%8F
概率期望学习:http://kicd.blog.163.com/blog/static/126961911200910168335852/
ACcode:
</pre><pre name="code" class="cpp">#include<cstdio> #include<iostream> #include<cmath> #include<algorithm> #include<cstring> using namespace std; int n,m; //N(1≤N≤100000) and M(0≤M≤1000). #define maxN 100001 #define maxM 1001 int jump[maxN]; double dp[maxN]; int main() { int a,b; while(scanf("%d%d",&n,&m)==2,n) { memset(dp,0,sizeof(dp)); memset(jump,-1,sizeof(jump)); for(int i=1;i<=m;i++) { scanf("%d%d",&a,&b); jump[a]=b; } for(int i=n-1;i>=0;i--) { if(jump[i]!=-1) { dp[i]=dp[jump[i]]; } else { for(int k=1;k<=6;k++) { if(i+k>n)break; dp[i]+=dp[i+k]; } dp[i]/=6; dp[i]+=1; } } printf("%.4lf\n",dp[0]); } return 0; }
相关文章推荐
- sgu495
- 高校联合Rating
- poj_2151
- HDU 5245 Joyful (数学概率求期望)
- Good Bye 2014 CF 500D
- Codeforces 148D Bag of mice(概率dp)
- poj 3071Football(概率dp)
- SGU 495 Kids and Prizes (概率DP)
- UVA11762 Race to 1
- UVA11722Joining with Friend概率dp
- hdu4336Card Collector_概率dp
- hdu3853LOOPS概率dp
- CodeForces 518D - R2D2 and Droid Army(概率dp)
- hdu 4418 高斯消元+概率dp
- bzoj3191: [JLOI2013]卡牌游戏
- CodeForces 261B Maxim and Restaurant
- HDU 4405 Aeroplane chess(概率DP)
- HDU 3853 LOOPS(概率DP)
- POJ 2151 Check the difficulty of problems(概率DP)
- Codeforces Round #105 (Div. 2) (D. Bag of mice(概率DP))