The number of steps
Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^
题目描述
Mary stands in a strange maze, the maze looks like a triangle（the first layer have one room，the second layer have two rooms，the third layer have three
rooms …）. Now she stands at the top point(the first layer), and the KEY of this maze is in the lowest layer’s leftmost room. Known that each room can only access to its left room and lower left and lower right rooms .If a room doesn’t have its left room, the
probability of going to the lower left room and lower right room are a and b (a + b = 1 ). If a room only has it’s left room, the probability of going to the room is 1. If a room has its lower left, lower right rooms and its left room, the probability of going
to each room are c, d, e (c + d + e = 1). Now , Mary wants to know how many steps she needs to reach the KEY. Dear friend, can you tell Mary the expected number of steps required to reach the KEY?
输入
输出
示例输入
3 0.3 0.7 0.1 0.3 0.6 0
示例输出
3.41
提示
来源
示例程序
水题概率dp
/*************************************************************************
> File Name: sdut2623.cpp
> Author: ALex
> Mail: 405045132@qq.com
> Created Time: 2014年12月28日 星期日 15时45分25秒
************************************************************************/
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
double a, b, c, d, e;
double dp[1000][1000];
int n;
double dfs(int i, int j)
{
if (i > n || i < 1 || j < 1 || j > i)
{
return 0;
}
if (dp[i][j] != -1)
{
return dp[i][j];
}
if (j <= 1) //没有左边的房间
{
dp[i][j] = a * dfs(i + 1, j) + b * dfs(i + 1, j + 1) + 1;
}
else
{
if (i == n)
{
dp[i][j] = dfs(i, j - 1) + 1;
}
else
{
dp[i][j] = e * dfs(i, j - 1) + c * dfs(i + 1, j) + d * dfs(i + 1, j + 1) + 1;
}
}
// printf("dp[%d][%d] = %.2f\n", i, j, dp[i][j]);
return dp[i][j];
}
int main()
{
while (~scanf("%d", &n), n)
{
for (int i = 0; i <= n; ++i)
{
for (int j = 0; j <= n; ++j)
{
dp[i][j] = -1;
}
}
dp[n][1] = 0;
scanf("%lf%lf%lf%lf%lf", &a, &b, &c, &d, &e);
printf("%.2f\n", dfs(1, 1));
}
return 0;
}