hdu4576Robot

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#hdu #概率dp #动态规划

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Robot

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1783 Accepted Submission(s): 655

Problem 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

Sample Output

      
       0.5000
0.2500

Source

2013ACM-ICPC杭州赛区全国邀请赛

Statistic | Submit | Discuss | Note

做这道题时我很无语，因为是挑一道最简单的题来做，但是自己真的很菜，很水！！

明知道这道题是水题都是在看到人家的结题报告才知道这是概率dp。

对自己的水平很失望。。

还有我看到dp的循环次数时以为会超时（一百万*2*200）这样都不超时，Orz~~Orz~~Orz~~Orz~~Orz！！

用b数组代表是这个状态，a数组代表是上一个状态然后有

b[(j-w+n)%n]+= a[j]*0.5;
b[(j+w)%n]+= a[j]*0.5;

#include<cstdio>
#include<cstring>
double a[212],b[212];
int main()
{
    int n,m,l,r,w,j;
    double va;
    while(scanf("%d%d%d%d",&n,&m,&l,&r))
    {
        if(n+m+l+r==0)
        break;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        a[0] = 1.0;
        for(int i=0;i<m;i++)
        {
            scanf("%d",&w);
            for(j=0;j<n;j++)
            {
                if(a[j]>0)//记住一定要这个剪枝，不然就很容易超时 
                {
                va = a[j]*0.5;
                b[(j-w+n)%n]+= va;
                b[(j+w)%n]+= va;
                }    
            }  
            for(j=0;j<n;j++)
            {
                a[j] = b[j];
                b[j] = 0;   //还有救市这个一定要初始化为0，不然就会错。 
            }      
            
        }    
        l--;
        r--;
        double ans = 0.0;
        for(;l<=r;l++)
        ans+=a[l];
        printf("%0.4f\n",ans);
    }    
    return 0;
}