HDU 4405 Aeroplane chess

Aeroplane chess

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4696    Accepted Submission(s): 2954

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
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 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 0 8 3 2 4 4 5 7 8 0 0

 

Sample Output

1.1667 2.3441

 

Source

2012 ACM/ICPC Asia Regional Jinhua Online

 

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#include
#include
#include
#include
#define N 100005
using namespace std;
double dp[N];
int Nxt[N];

inline void read(int &x) {
    x = 0; register char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)) x = x * 10 + c - '0',c = getchar();
}

int main(int argc,char *argv[]) {
    int n,m;
    while(scanf("%d %d",&n,&m) != EOF) {
        if(n == 0 && m == 0) break;
        memset(Nxt,-1,sizeof Nxt );
        for(int u,v,i=1; i<=m; ++i) {
            scanf("%d %d",&u,&v);
            Nxt[u] = v;
        }
        memset(dp,0,sizeof dp );
        for(int i=n-1; i>=0; --i) {
            if(Nxt[i] == -1){
                for(int j=1; j<=6; ++j)
                    dp[i] += dp[i + j] / 6.0;
                dp[i] += 1;
            }
            else dp[i] = dp[Nxt[i]];
        }
        printf("%.4lf\n",dp[0]);
    }
    
    return 0;
}