HDU - 3853(简单概率dp ， 注意边界)

本题目直接按题目意思推方程转移就行了，但是有个坑点，就是对于一个点i,j从1,1开始不可达，那么该点的p1（转移到自己）就有可能为1，所以当转移到一个点的概率为0的时候

那么就不往该点转移。

//#pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <set>#include <map>#include <string>#include <list>#include <cstdlib>#include <queue>#include <stack>#include <cmath>#define ALL(a) a.begin(), a.end()#define clr(a, x) memset(a, x, sizeof a)#define fst first#define snd second#define pb push_back#define lowbit(x) (x&(-x))#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define rep1(i,x,y) for(int i=x;i<=y;i++)#define rep(i,n) for(int i=0;i<(int)n;i++)using namespace std;const double eps = 1e-10;typedef long long LL;typedef long long ll;typedef pair<int, int> pii;const int inf =0x3f3f3f3f;const int N = 1010;double p[N][N][4],E[N][N];bool vis[N][N];int n,m;double dp(int i,int j){   if(vis[i][j]) return E[i][j];   vis[i][j] = 1;   if(i == n && j == m) return E[i][j] = 0;   E[i][j]=2;   if(j < m && p[i][j][2] > eps) E[i][j]+=dp(i,j+1)*p[i][j][2];   if(i < n && p[i][j][3] > eps) E[i][j]+=dp(i+1,j)*p[i][j][3];   E[i][j]/=(1-p[i][j][1]);   return E[i][j];}int main(){   while(scanf("%d %d",&n,&m)==2){      rep1(i,1,n) rep1(j,1,m)        scanf("%lf %lf %lf",&p[i][j][1],&p[i][j][2],&p[i][j][3]);      memset(vis,false,sizeof(vis));      printf("%.3lf\n",dp(1,1));   }   return 0;}