HDU 3853 LOOPS 概率DP 期望

题意：在$(i,j)$处的点有$rate_{(i,j,1)}$的概率留在$(i,j)$处，有$rate_{(i,j,2)}$的概率走向$(i,j+1)$处,有$rate_{(i,j,3)}$的概率走向$(i+1,j)$处，走一步需要花费2，问从$(1,1)$走到$(n,m)$处花费的期望是多少?

思路：期望是概率的反推过程

$$E_{ij} = rate_{(i,j,1)} * E_{ij} + rate_{(i,j,2)} * E_{i,j+1} + rate_{(i,j,3)} * E_{i+1,j} + 2$$

$$+2 是走一步的代价 是从之前那个状态走一步才能到E_{ij}$$

$$移项后$$

$$E_{ij} = (rate_{(i,j,2)} * E_{i,j+1} + rate_{(i,j,3)} * E_{i+1,j} + 2 )/ (1-rate_{(i,j,1)})$$

http://acm.hdu.edu.cn/showproblem.php?pid=3853

/*********************************************
    Problem : HDU 3853
    Author  : NMfloat
    InkTime (c) NM . All Rights Reserved .
********************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>

#define rep(i,a,b)  for(int i = (a) ; i <= (b) ; i ++) //遍历
#define rrep(i,a,b) for(int i = (b) ; i >= (a) ; i --) //反向遍历
#define repS(it,p) for(auto it = p.begin() ; it != p.end() ; it ++) //遍历一个STL容器
#define repE(p,u) for(Edge * p = G[u].first ; p ; p = p -> next) //遍历u所连接的点
#define cls(a,x)   memset(a,x,sizeof(a))
#define eps 1e-8

using namespace std;

const int MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5+5;
const int MAXE = 2e5+5;

typedef long long LL;
typedef unsigned long long ULL;

int T,n,m;

int fx[] = {0,1,-1,0,0};
int fy[] = {0,0,0,-1,1};

double rate[1005][1005][4];
double E[1005][1005];
int v[1005][1005];
//v[i][j]存这一点rate[i][j][1]是否是1,因为如果rate[i][j][1]是1的话,这点就永远走不出去了。(n,m)是特殊情况,因为这点是终点。

void input() {
    rep(i,1,n) rep(j,1,m) rep(k,1,3) scanf("%lf",&rate[i][j][k]);
}

void solve() {
    //E[i][j] = rate[i][j][1] * E[i][j] + rate[i][j][2] * E[i][j+1] + rate[i][j][3] * E[i+1][j] + 2;
    //移项
    // +2 是走一步的代价  是从之前那个状态走一步才能到E[i][j];
    //E[i][j] =  (rate[i][j][2] * E[i][j+1] + rate[i][j][3] * E[i+1][j] + 2) / (1 - rate[i][j][1]);
    cls(v,0);
    rep(i,1,n) rep(j,1,m) if(fabs(rate[i][j][2]+rate[i][j][3]) > eps) v[i][j] = 1;
    v[n][m] = 1;
    rrep(i,1,n) rrep(j,1,m) { 
        E[i][j] = 0;
        if(i == n && j == m) continue;
        if(v[i][j+1]) {
            E[i][j] +=  (rate[i][j][2] * E[i][j+1]) / (1-rate[i][j][1]);
        }
        if(v[i+1][j]) {
            E[i][j] +=  (rate[i][j][3] * E[i+1][j]) / (1-rate[i][j][1]);
        }
        E[i][j] +=  2 / (1-rate[i][j][1]);
    }
    printf("%.3f\n",E[1][1]);
}

int main(void) {
    //freopen("a.in","r",stdin);
    while(~scanf("%d %d",&n,&m)) {
        input();
        solve();
    }
    return 0;
}