HDU 4405 Aeroplane chess 概率dp
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Aeroplane chess
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1961 Accepted Submission(s): 1294
Problem Description
Hzz loves aeroplane chess very much. The chess map contains N+1 grids labeled from 0 to N. Hzz starts at grid 0. For each step he throws a dice(a dice have six faces with equal probability to face up and the numbers on the faces are 1,2,3,4,5,6). When Hzz is at grid i and the dice number is x, he will moves to grid i+x. Hzz finishes the game when i+x is equal to or greater than N.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.
Please help Hzz calculate the expected dice throwing times to finish the game.
There are also M flight lines on the chess map. The i-th flight line can help Hzz fly from grid Xi to Yi (0<Xi<Yi<=N) without throwing the dice. If there is another flight line from Yi, Hzz can take the flight line continuously. It is granted that there is no two or more flight lines start from the same grid.
Please help Hzz calculate the expected dice throwing times to finish the game.
Input
There are multiple test cases.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
Each test case contains several lines.
The first line contains two integers N(1≤N≤100000) and M(0≤M≤1000).
Then M lines follow, each line contains two integers Xi,Yi(1≤Xi<Yi≤N).
The input end with N=0, M=0.
Output
For each test case in the input, you should output a line indicating the expected dice throwing times. Output should be rounded to 4 digits after decimal point.
Sample Input
2 08 32 44 57 80 0
Sample Output
1.16672.3441
Source
2012 ACM/ICPC Asia Regional Jinhua Online
概率dp,参考了一些人的代码,想明白一些问题,,
/** Author: ☆·aosaki(*’(OO)’*) niconiconi★ **///#pragma comment(linker, "/STACK:1024000000,1024000000")//#include<bits/stdc++.h>#include <iostream>#include <sstream>#include <cstdio>#include <cstring>#include <algorithm>#include <functional>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <list>#include <stack>//#include <tuple>#define mem(a) memset(a,0,sizeof(a))#define mem1(a) memset(a,-1,sizeof(a))#define lp(k,a) for(int k=1;k<=a;k++)#define lp0(k,a) for(int k=0;k<a;k++)#define lpn(k,n,a) for(int k=n;k<=a;k++)#define lpd(k,n,a) for(int k=n;k>=a;k--)#define sc(a) scanf("%d",&a)#define sc2(a,b) scanf("%d %d",&a,&b)#define lowbit(x) (x&(-x))#define ll long long#define pi pair<int,int>#define vi vector<int>#define PI acos(-1.0)#define pb(a) push_back(a)#define mp(a,b) make_pair(a,b)#define TT cout<<"*****"<<endl;#define TTT cout<<"********"<<endl;inline int gcd(int a,int b){ return a==0?b:gcd(b%a,a);}#define INF 1e9#define eps 1e-8#define mod 10007#define MAX 10010using namespace std;int n,m,a,b;int x[1000010];double dp[1000010];int main(){ //freopen("in.txt","r",stdin); while(~sc2(n,m)) { if(n==0 && m==0) return 0; mem(x); lp(i,m) { sc2(a,b); x[a]=b; } mem(dp); dp[n]=0; lpd(i,n-1,0) { if(x[i]) dp[i]=dp[x[i]]; else { lpn(j,1,6) dp[i]+=dp[i+j]/6.0; dp[i]+=1; } } printf("%.4f\n",dp[0]); } return 0;}
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