1151 - Snakes and Ladders

Time Limit: 2 second(s)    Memory Limit: 32 MB

'Snakes and Ladders' or 'Shap-Ludu' is a game commonly played in Bangladesh. The game is so common that it would be tough to find a person who hasn't played it. But those who haven't played it (unlucky of course!) the rules are as follows. There is a 10 x 10 board containing some cells numbered from 1 to 100.

  1. You start at position 1.
  2. Each time you throw a perfect dice containing numbers 1 to 6.
  3. There are some snakes and some ladders in the board. Ladders will take you up from one cell to another. Snakes will take you down.
  4. If you reach a cell that contains the bottom part of a ladder, you will immediately move to the cell which contains the upper side of that ladder. Similarly if you reach a cell that has a snake-head you immediately go down to the cell where the tail of that snake ends.
  5. The board is designed so that from any cell you can jump at most once. (For example there is a snake from 62 to 19, assume that another is from 19 to 2. So, if you reach 62, you will first jump to 19, you will jump to 2. These kinds of cases will not be given)
  6. There is no snake head in the 100-th cell and no ladder (bottom part) in the first cell.
  7. If you reach cell 100, the game ends. But if you have to go outside the board in any time your move will be lost. That means you will not take that move and you have to throw the dice again.

Now given a board, you have to find the expected number of times you need to throw the dice to win the game. The cases will be given such that a result will be found.

Input

Input starts with an integer T (≤ 105), denoting the number of test cases.

The first line of a case is a blank line. The next line gives you an integer n denoting the number of snakes and ladders. Each of the next n lines contain two integers a and b (1 ≤ a, b ≤ 100, a ≠ b). If a < b, it means that there is a ladder which takes you from a to b. If a > b, it means that there is a snake which takes you from a to b. Assume that the given board follows the above restrictions.

Output

For each case of input, print the case number and the expected number of times you need to throw the dice. Errors less than 10-6 will be ignored.

Sample Input

2

14

4 42

9 30

16 8

14 77

32 12

37 58

47 26

48 73

62 19

70 89

71 67

80 98

87 24

96 76

0

Output for Sample Input

Case 1: 31.54880806

Case 2: 33.0476190476

主要题意就不解释了。。

我们设从点i到100的步数期望为Ei。

则:

如果Ei有连向其他格子的边,设走到to[i],则Ei=Etoi。

否则Ei=(Ex1+Ex2+...+Exk)*(1/6)+1。其中,k=min(6,100-i),x1+1=x2,x2+1=x3,......xi+1=xi+1。

但是我们发现,to[i]可能大于i,也可能小于i,所以不能直接DP或递推。

所以相当于解一个有100个100元方程的方程组。其中最后一个方程已经确定,且得到E[100]=0。

那么,就相当于用高斯消元解一个有唯一解的实数方程组了。

code:

 #include<bits/stdc++.h>
 #define Ms(a,x) memset(a,x,sizeof a)
 using namespace std;
 ;
 int n,got[N]; double a[N][N],E[N];
 ?x:-x;}
 void Gauss(int equ,int var) {
     ,col=,cho;
     for (; row<=equ&&col<=var; row++,col++) {
         cho=row;
         ; i<=equ; i++)
             if (abso(a[i][col])>abso(a[cho][col])) cho=col;
         if (cho!=row)
             ; i++) swap(a[cho][i],a[row][i]);
         ) {col--; continue;}
         ; i<=equ; i++) ) {
             double k=a[i][col]/a[row][col];
             ; j++) a[i][j]-=k*a[row][j];
         }
     }
     for (int i=var; i; i--) {
         ];
         ; j<=var; j++) re-=a[i][j]*E[j];
         E[i]=re/a[i][i];
     }
 }
 int main() {
     int T; scanf("%d",&T);
     ; ts<=T; ts++) {
         cin>>n,Ms(got,),Ms(a,),Ms(E,);
         ,x,y; i<=n; i++)
             scanf("%d%d",&x,&y),got[x]=y;
         ,c; i<; i++) if (!got[i]) {
             c=min(,-i),a[i][i]=c,a[i][]=;
             ; j<=&&i+j<=; j++) a[i][i+j]=-;
         } ,a[i][got[i]]=-,a[i][]=;
         a[][]=,a[][]=;
         Gauss(,);
         printf(]);
     }
     ;
 }

[lightoj P1151] Snakes and Ladders的更多相关文章

  1. LightOJ - 1151 Snakes and Ladders —— 期望、高斯消元法

    题目链接:https://vjudge.net/problem/LightOJ-1151 1151 - Snakes and Ladders    PDF (English) Statistics F ...

  2. LightOJ 1151 - Snakes and Ladders 高斯消元+概率DP

    首先来个期望的论文,讲的非常好,里面也提到了使用线性方程组求解,尤其适用于有向图的期望问题. 算法合集之<浅析竞赛中一类数学期望问题的解决方法> http://www.lightoj.co ...

  3. LightOJ - 1151 Snakes and Ladders

    LightOJ - 1151 思路: 将期望dp[x]看成自变量,那么递推式就可以看成方程组,用高斯消元求方程组的解就能求解出期望值 高斯消元求解的过程也是期望逆推的过程,注意边界情况的常数项,是6/ ...

  4. LightOJ 1151 Snakes and Ladders(概率DP + 高斯消元)

    题意:1~100的格子,有n个传送阵,一个把进入i的人瞬间传送到tp[i](可能传送到前面,也可能是后面),已知传送阵终点不会有另一个传送阵,1和100都不会有传送阵.每次走都需要掷一次骰子(1~6且 ...

  5. LightOJ 1151 Snakes and Ladders 期望dp+高斯消元

    题目传送门 题目大意:10*10的地图,不过可以直接看成1*100的,从1出发,要到达100,每次走的步数用一个大小为6的骰子决定.地图上有很多个通道 A可以直接到B,不过A和B大小不确定   而且 ...

  6. LightOJ - 1151 Snakes and Ladders(概率dp+高斯消元)

    有100个格子,从1开始走,每次抛骰子走1~6,若抛出的点数导致走出了100以外,则重新抛一次.有n个格子会单向传送到其他格子,G[i]表示从i传送到G[i].1和100不会有传送,一个格子也不会有两 ...

  7. Snakes and Ladders LightOJ - 1151( 概率dp+高斯消元)

    Snakes and Ladders LightOJ - 1151 题意: 有100个格子,从1开始走,每次抛骰子走1~6,若抛出的点数导致走出了100以外,则重新抛一次.有n个格子会单向传送到其他格 ...

  8. [Swift]LeetCode909. 蛇梯棋 | Snakes and Ladders

    On an N x N board, the numbers from 1 to N*N are written boustrophedonically starting from the botto ...

  9. light oj 1151 - Snakes and Ladders 高斯消元+概率DP

    思路: 在没有梯子与蛇的时候很容易想到如下公式: dp[i]=1+(∑dp[i+j])/6 但是现在有梯子和蛇也是一样的,初始化p[i]=i; 当有梯子或蛇时转移为p[a]=b; 这样方程变为: dp ...

随机推荐

  1. 用python写多线程

    import threading #首先导入threading 模块,这是使用多线程的前提 from time import ctime,sleep def music(func): ): print ...

  2. JavaScript 数组插入元素并排序

    1.插入类排序 插入类排序的思想是:在一个已排好序的序列区内,对待排序的无序序列中的记录逐个进行处理,每一步都讲待排序的记录和已排好的序列中的记录进行比较,然后有序的插入到该序列中,直到所有待排序的记 ...

  3. gitlab重置root的密码

    环境:gitlab 忘记了root密码,无法登陆gitlab 解决: gitlab-ctl start 保证gitlab处于启动状态,&保证redis处于启动状态 gitlab-rails c ...

  4. Mysql授权root用户远程登录

    默认情况下Mysql的root用户不支持远程登录,使用以下命令授权   [Charles@localhost ~]$ mysql -uroot -p123 MariaDB [(none)]> u ...

  5. sql server 操作列

    新增一列 ) 修改列类型: ) 修改列的名称: EXEC sp_rename 'tableName.column1' , 'column2' (把表名为tableName的column1列名修改为co ...

  6. uvm设计分析——reg

    项目中的reg_model一般只有一份,set到reg_sequence上,所以多个sequence并行启动结束的时候,reg model会成为一个共享资源. uvm_reg_field中的volat ...

  7. 用mpvue构建微信小程序

    背景 由于机器人协会进行鼓励大家多读书的活动,所以为了可以更好的.更有效果,所以我跟会长提了一个建议,做一个微信小程序,那么为什么是微信小程序呢? 1.现在微信小程序比较好,用户也比较多:利用微信小程 ...

  8. Oracle表空间迁移Move Tablespace

    move一个表到另外一个表空间时,索引不会跟着一起move,而且会失效.(LOB类型例外) move分为: *普通表move *分区表move *LONG,LOB大字段类型move来进行测试和说明. ...

  9. mysql的一些指令

    说起来mysql,mysql其实是特别简单操作的数据库.它有一下几方面的特点,亦或者是说优点,1,支持多语言:2.移植性比较好:3,我觉得最重要的一点就是它是开源的,因为开源,所以使用广泛,4.效率比 ...

  10. 2.0JAVA基础复习——JAVA语言的基础组成关键字和标识符

    JAVA语言的基础组成有: 1.关键字:被赋予特殊含义的单词. 2.标识符:用来标识的符号. 3.注释:用来注释说明程序的文字. 4.常量和变量:内存存储区域的表示. 5.运算符:程序中用来运算的符号 ...