HDU-3853 LOOPS 概率DP
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3853
简单概率DP,转移方程:f[i][j]=f[i][j]*p1+f[i][j+1]*p2+f[i+1][j]*p3+2 —> f[i][j]=(f[i][j+1]*p2+f[i+1][j]*p3+2)/(1-p1).
1 //STATUS:C++_AC_2828MS_32180KB 2 #include <functional> 3 #include <algorithm> 4 #include <iostream> 5 //#include <ext/rope> 6 #include <fstream> 7 #include <sstream> 8 #include <iomanip> 9 #include <numeric> 10 #include <cstring> 11 #include <cassert> 12 #include <cstdio> 13 #include <string> 14 #include <vector> 15 #include <bitset> 16 #include <queue> 17 #include <stack> 18 #include <cmath> 19 #include <ctime> 20 #include <list> 21 #include <set> 22 #include <map> 23 using namespace std; 24 //#pragma comment(linker,"/STACK:102400000,102400000") 25 //using namespace __gnu_cxx; 26 //define 27 #define pii pair<int,int> 28 #define mem(a,b) memset(a,b,sizeof(a)) 29 #define lson l,mid,rt<<1 30 #define rson mid+1,r,rt<<1|1 31 #define PI acos(-1.0) 32 //typedef 33 typedef __int64 LL; 34 typedef unsigned __int64 ULL; 35 //const 36 const int N=1010; 37 const int INF=0x3f3f3f3f; 38 const int MOD= 1000000007,STA=8000010; 39 const LL LNF=1LL<<55; 40 const double EPS=1e-9; 41 const double OO=1e30; 42 const int dx[4]={-1,0,1,0}; 43 const int dy[4]={0,1,0,-1}; 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31}; 45 //Daily Use ... 46 inline int sign(double x){return (x>EPS)-(x<-EPS);} 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;} 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;} 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;} 50 template<class T> inline T Min(T a,T b){return a<b?a:b;} 51 template<class T> inline T Max(T a,T b){return a>b?a:b;} 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);} 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);} 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));} 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));} 56 //End 57 58 double f[N][N],p[N][N][3]; 59 int n,m; 60 61 double dfs(int x,int y) 62 { 63 if(f[x][y]>=0)return f[x][y]; 64 if(x==n-1 && y==m-1)return f[x][y]=0; 65 f[x][y]=2/(1-p[x][y][0]); 66 if(p[x][y][1]>0) 67 f[x][y]+=dfs(x,y+1)*p[x][y][1]/(1-p[x][y][0]); 68 if(p[x][y][2]>0) 69 f[x][y]+=dfs(x+1,y)*p[x][y][2]/(1-p[x][y][0]); 70 return f[x][y]; 71 } 72 73 int main(){ 74 // freopen("in.txt","r",stdin); 75 int i,j; 76 while(~scanf("%d%d",&n,&m)) 77 { 78 mem(p,0); 79 for(i=0;i<n;i++){ 80 for(j=0;j<m;j++){ 81 scanf("%lf%lf%lf",&p[i][j][0],&p[i][j][1],&p[i][j][2]); 82 } 83 } 84 85 for(i=0;i<n;i++) 86 for(j=0;j<m;j++) 87 f[i][j]=-1; 88 dfs(0,0); 89 90 printf("%.3lf\n",f[0][0]); 91 } 92 return 0; 93 }