HDU 3853 LOOPS [简单概率DP]

最新推荐文章于 2025-04-11 15:47:45 发布

GrassTreeFlower

最新推荐文章于 2025-04-11 15:47:45 发布

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分类专栏：概率DP/概率/期望文章标签： math ACM 期望 HDUOJ acm-icpc

本文链接：https://blog.csdn.net/GrassTreeFlower/article/details/50176639

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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.

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.

题意：给出一个矩阵，每个点每一步有P0的概率在原点，P1的概率向右走，P2的概率向左走，问从左上角走到右下角的期望步数。

解法：很简单很明显的概率DP，DP[I][J]表示当前在I,J，走到右下角的期望步数，那么转移方程为：

DP[I][J]=(DP[I][J+1]*P1+DP[I+1][J]*P2)/(1-P0)

注意1-P0为0的时候，会是0/0的情况，应该返回0，否则计算的时候会出错，具体坑点看代码。

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<stdlib.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<bitset>
#pragma comment(linker, "/STACK:1024000000,1024000000")
template <class T>
bool scanff(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
#define inf 1073741823
#define llinf 4611686018427387903LL
#define PI acos(-1.0)
#define lth (th<<1)
#define rth (th<<1|1)
#define rep(i,a,b) for(int i=int(a);i<=int(b);i++)
#define drep(i,a,b) for(int i=int(a);i>=int(b);i--)
#define gson(i,root) for(int i=ptx[root];~i;i=ed[i].next)
#define tdata int testnum;scanff(testnum);for(int cas=1;cas<=testnum;cas++)
#define mem(x,val) memset(x,val,sizeof(x))
#define mkp(a,b) make_pair(a,b)
#define findx(x) lower_bound(b+1,b+1+bn,x)-b
#define pb(x) push_back(x)
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

double p[1010][1010][3];
double dp[1010][1010];
bool vis[1010][1010];
int n,m;
double dfs(int i,int j){
    if(j>m||i>n||i==n&&j==m)return 0;
    if(vis[i][j])return dp[i][j];
    vis[i][j]=1;
    dp[i][j]=2.0;
    //必须保证有概率走过去，否则那些点肯定不用走
    //那些点有可能一直在本身循环，导致答案错误，产生0/0的情况
    if(p[i][j][1]>0.0)dp[i][j]+=p[i][j][1]*dfs(i,j+1);
    if(p[i][j][2]>0.0)dp[i][j]+=p[i][j][2]*dfs(i+1,j);
    dp[i][j]/=(1.0-p[i][j][0]);
    return dp[i][j];
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        rep(i,1,n)rep(j,1,m){
            vis[i][j]=0;
            scanff(p[i][j][0]);
            scanff(p[i][j][1]);
            scanff(p[i][j][2]);
        }
        double ans=dfs(1,1);
        printf("%.3f\n",ans);
    }
}