标签：des style blog http color io os ar for
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?
3 0.3 0.7 0.1 0.3 0.6 0
3.41
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
double dp[100][100] ;
double a , b , c , d , e ;
int i , j , n ;
int ff(int x,int y)
{
if( x <= n && y >=(n+1)-x )
return 1 ;
return 0 ;
}
void f()
{
return ;
}
int main()
{
while(scanf("%d", &n) && n)
{
scanf("%lf %lf", &a, &b);
scanf("%lf %lf %lf", &c, &d, &e);
memset(dp,0,sizeof(dp));
for(i = n ; i >= 1 ; i--)
{
for(j = (n+1)-i ; j <= n ; j++)
{
if(i == n && j == (n+1)-i) continue ;
else if( i == n )
dp[i][j] = 1.0*( dp[i][j-1] ) + 1.0 ;
else
{
if( j == (n+1)-i )
dp[i][j] = a*dp[i+1][j-1] + b*dp[i+1][j] + 1.0 ;
else
dp[i][j] = c*dp[i+1][j-1] + d*dp[i+1][j] + e*dp[i][j-1] + 1.0 ;
}
}
}
printf("%.2lf\n", dp[1][n]);
}
return 0;
}
sdut2623--The number of steps（概率dp第一弹）
标签：des style blog http color io os ar for
原文地址：http://blog.csdn.net/winddreams/article/details/40381229
