hdu4487 Maximum Random Walk

Maximum Random Walk

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 128

Problem DescriptionConsider the classic random walk: at each step, you have a 1/2 chance of taking a step to the left and a 1/2 chance of taking a step to the right. Your expected position after a period of time is zero; that is, the average over many
such random walks is that you end up where you started. A more interesting question is what is the expected rightmost position you will attain during the walk.

InputThe first line of input contains a single integer P, (1 <= P <= 15), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of a single line of input consisting of four space-separated values. The first value is an integer K, which is the data set number. Next is an integer n, which is the number of steps to take (1 <= n <= 100). The final two are double precision
floating-point values L and R
which are the probabilities of taking a step left or right respectively at each step (0 <= L <= 1, 0 <= R <= 1, 0 <= L+R <= 1). Note: the probably of not taking a step would be 1-L-R.

OutputFor each data set there is a single line of output. It contains the data set number, followed by a single space which is then followed by the expected (average) rightmost position you will obtain during the walk, as a double precision
floating point value to four decimal places.

Sample Input

31 1 0.5 0.52 4 0.5 0.53 10 0.5 0.4

Sample Output

1 0.50002 1.18753 1.4965

SourceGreater New York 2012

Recommendliuyiding機率dp,　我們可以發現，dp[i][j][k]表示走了i步，到達了j,最遠到過k,　我們發現，k是一定要大於j的，否則為0,因為我們要求的是走到最右端的期望，也就是要求走到最遠右邊每一個點的距離*它的機率，這樣，我們可以先把機率算出來，我們可以得到壯態轉移方程，如果，j>k那麼就是 dp[i][j][k]=dp[i-1][j-1][k]*r+dp[i-1][j-1][k-1]*r+dp[i-1][j][k]*re;否則，就是                        dp[i][j][k]=dp[i-1][j+1][k]*l+dp[i-1][j][k]*re+dp[i-1][j-1][k]*r;這樣，馬上，就可以算出來了，注意這裡因為有左右，所以要標記一個原點，這樣就好處理了！

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;double dp[225][225][225];int fmax(int a,int b){    if(a>b)    return a;    return b;}int main(){    int tcase,t,n,i,j,k;    double l,r,re;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%d%d%lf%lf",&t,&n,&l,&r);        re=1-l-r;        dp[0][110][110]=1;        for(i=1;i<=n;i++)        {            for(j=110-i;j<=i+110;j++)            {                for(k=fmax(j,110);k<=110+i;k++)                {                    if(j==k)                        dp[i][j][k]=dp[i-1][j-1][k]*r+dp[i-1][j-1][k-1]*r+dp[i-1][j][k]*re;                    else                        dp[i][j][k]=dp[i-1][j+1][k]*l+dp[i-1][j][k]*re+dp[i-1][j-1][k]*r;                    //printf("%.4f\n",dp[i][j][k]);                }            }        }        double ans=0;        for(i=110-n;i<=110+n;i++)        {            for(k=110;k<=110+n;k++)            {                ans+=dp[n][i][k]*(k-110);            }        }        printf("%d %.4f\n",t,ans);    }    return 0;}