首页 > ACM题库 > HDU-杭电 > HDU 4576-Robot-动态规划-[解题报告]HOJ
2015
09-16

HDU 4576-Robot-动态规划-[解题报告]HOJ

Robot

问题描述 :

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.

Changsha Marathon

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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 <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 220;
const double eps = 0.0;
double dp[N][2];

int main(){
    int n, m, l, r;    
    while(cin >> n >> m >> l >> r){
        if(n == 0 && m == 0 && l == 0 && r == 0)
            break;
        int w, i, j;
        for(i = 0;i < n;i++)
            dp[i][0] = dp[i][1] = eps;
        dp[0][0] = 1.0;
        for(j = 1;j <= m;j++){
            scanf("%d", &w);
            for(i = 0;i < n;i++){
                dp[(i + w) % n][j&1] += dp[i][!(j&1)] * 0.5;
                int x = i - w;
                while(x < 0)
                	x += n;
                 dp[x][j&1] += dp[i][!(j&1)] * 0.5;    
            }
            for(i = 0;i < n;i++)
                dp[i][!(j&1)] = eps;
        }
        double res = 0.0;
        l--, r--;
        for(i = l;i <= r;i++)
            res += dp[i][m&1];
        printf("%.4lf\n", res);
    }    
    return 0;
}

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参考:http://blog.csdn.net/joy_go/article/details/9882071