hdu 3853 概率dp

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LOOPS

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 3738 Accepted Submission(s): 1488

Problem Description

Akemi Homura is a Mahou Shoujo (Puella Magi/Magical Girl).

Homura wants to help her friend Madoka save the world. But because of the plot of the Boss Incubator, she is trapped in a labyrinth called LOOPS.
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The planform of the LOOPS is a rectangle of R*C grids. There is a portal in each grid except the exit grid. It costs Homura 2 magic power to use a portal once. The portal in a grid G(r, c) will send Homura to the grid below G (grid(r+1, c)), the grid on the right of G (grid(r, c+1)), or even G itself at respective probability (How evil the Boss Incubator is)!
At the beginning Homura is in the top left corner of the LOOPS ((1, 1)), and the exit of the labyrinth is in the bottom right corner ((R, C)). Given the probability of transmissions of each portal, your task is help poor Homura calculate the EXPECT magic power she need to escape from the LOOPS.

Input

The first line contains two integers R and C (2 <= R, C <= 1000).

The following R lines, each contains C*3 real numbers, at 2 decimal places. Every three numbers make a group. The first, second and third number of the cth group of line r represent the probability of transportation to grid (r, c), grid (r, c+1), grid (r+1, c) of the portal in grid (r, c) respectively. Two groups of numbers are separated by 4 spaces.

It is ensured that the sum of three numbers in each group is 1, and the second numbers of the rightmost groups are 0 (as there are no grids on the right of them) while the third numbers of the downmost groups are 0 (as there are no grids below them).

You may ignore the last three numbers of the input data. They are printed just for looking neat.

The answer is ensured no greater than 1000000.

Terminal at EOF

Output

A real number at 3 decimal places (round to), representing the expect magic power Homura need to escape from the LOOPS.

Sample Input

2 2 0.00 0.50 0.50 0.50 0.00 0.50 0.50 0.50 0.00 1.00 0.00 0.00

Sample Output

6.000

Source

2011 Invitational Contest Host by BUPT

题意：在N*M的图里，对于每个点，给你向下或者向右不动的概率，问你从0，0到n，m的期望

题解：dp[i][j]表示从（i，j）到(n,m)的期望大小

dp[i][j]=p2[i][j]*dp[i][j+1]/(1-p1[i][j])+p3[i][j]*dp[i+1][j]/(1-p1[i][j])+2/(1-p1[i][j]);

///1085422276
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<cmath>
#include<map>
#include<bitset>
#include<set>
#include<vector>
using namespace std ;
typedef long long ll;
#define push_back pb
#define mem(a) memset(a,0,sizeof(a))
#define TS printf("111111\n");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b) scanf("%d%d",&a,&b)
#define mod 1000000007
#define inf 100000
#define maxn 1001
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
//******************************************************************
int n,m;
double p1[maxn][maxn],p2[maxn][maxn],p3[maxn][maxn];
double dp[maxn][maxn];
int main()
{
    while(READ(n,m)!=EOF)
    {
           FOR(i,0,n-1)
           FOR(j,0,m-1)
               scanf("%lf%lf%lf",&p1[i][j],&p2[i][j],&p3[i][j]);
           mem(dp);
           FORJ(i,n-1,0)
           FORJ(j,m-1,0)
           {
              if(i==n-1&&j==m-1){continue;}
              if(p1[i][j]==1)dp[i][j]=0;
              else
              dp[i][j]=p2[i][j]*dp[i][j+1]/(1-p1[i][j])+p3[i][j]*dp[i+1][j]/(1-p1[i][j])+2/(1-p1[i][j]);
           }
         printf("%.3f\n",dp[0][0]);
    }
    return 0;
}

代码

hdu 3853 概率dp

原文：http://www.cnblogs.com/zxhl/p/4781635.html

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