HDU 4546 Robot(概率DP）

Robot Time Limit:4000MS     Memory Limit:102400KB     64bit IO Format:%I64d & %I64u Submit  Status

Description

Michael has a telecontrol robot. One day he put the robot on a loop with n cells. The cells are numbered from 1 to n clockwise. 

At first the robot is in cell 1. Then Michael uses a remote control to send m commands to the robot. A command will make the robot walk some distance. Unfortunately the direction part on the remote control is broken, so for every command the robot will chose a direction(clockwise or anticlockwise) randomly with equal possibility, and then walk w cells forward. 

Michael wants to know the possibility of the robot stopping in the cell that cell number >= l and <= r after m commands. 

Input

There are multiple test cases. 

Each test case contains several lines. 

The first line contains four integers: above mentioned n(1≤n≤200) ,m(0≤m≤1,000,000),l,r(1≤l≤r≤n). 

Then m lines follow, each representing a command. A command is a integer w(1≤w≤100) representing the cell length the robot will walk for this command.   

The input end with n=0,m=0,l=0,r=0. You should not process this test case. 

Output

For each test case in the input, you should output a line with the expected possibility. Output should be round to 4 digits after decimal points.

Sample Input

3 1 1 2
1
5 2 4 4
1
2
0 0 0 0      

Sample Output

0.5000
0.2500      

题意：杭州邀请赛的一道题；有标记着1~n的环，初始时有个机器人在1的位置，有m个操作，每个操作是使机器人随机的顺时针或逆时针的走w步，求所有操作过后机器人停在区间[l,r]内的概率；

思路：挺明显的概率dp，首先可以很直观的定义出状态dp[i][j]，表示第j次操作后停留在第i位置的概率，转移方程是

dp[(i + w) % n][j] = dp[i[j-1] * 0.5；

dp[i-w][j]=dp[i][j-1]*0.5;其中i-w不断的加n使它大于等于0；

但是题目操作数比较大，这样会超内存，观察发现每次转移只和上一次操作有关，所以可以用个滚动数组来节省空间；优化空间后转移方程是

dp[(i + w) % n][j&1] += dp[i][!(j&1)] * 0.5；

dp[x][j&1] += dp[i][!(j&1)] * 0.5；x=i-w，x也要叠加到非负；

最后答案就是所有操作完后，位置l到r的概率总和。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <vector>

using namespace std;

const int N = 2100;

const int inf = 0x3f3f3f3f;

double dp[N][2];

int main()

{

    int n, m, l, r;

    while(scanf("%d %d %d %d", &n, &m, &l, &r)!=EOF)

    {

        if(n==0 && m==0 && l==0 && r==0)

            break;

        for(int i=1;i<=n;i++)

            dp[i][1]=dp[i][0]=0;

        dp[1][0]=1;

        for(int i=1;i<=m;i++)

        {

            int k;

            scanf("%d", &k);

            for(int j=1;j<=n;j++)

            {

                int x=(j+k)%n, y=(j-k);

                while(y < 0)

                    y += n;

                if(x==0) x=n;

                if(y==0) y=n;

                dp[x][i&1]+=dp[j][!(i&1)]*0.5;

                dp[y][i&1]+=dp[j][!(i&1)]*0.5;

            }

            for(int j=1;j<=n;j++)

                dp[j][!(i&1)]=0;

        }

        double sum=0;

        for(int i=l;i<=r;i++)

        {

            sum+=dp[i][m&1];

        }

        printf("%.4f\n",sum);

    }

    return 0;

}