[Codeforces605E]Intergalaxy-Trips-概率与期望

Intergalaxy Trips

The scientists have recently discovered wormholes — objects in space that allow to travel very long distances between galaxies and star systems.

The scientists know that there are n galaxies within reach. You are in the galaxy number 1 and you need to get to the galaxy number n. To get from galaxy i to galaxy j, you need to fly onto a wormhole (i, j) and in exactly one galaxy day you will find yourself in galaxy j.

Unfortunately, the required wormhole is not always available. Every galaxy day they disappear and appear at random. However, the state of wormholes does not change within one galaxy day. A wormhole from galaxy i to galaxy j exists during each galaxy day taken separately with probability pij. You can always find out what wormholes exist at the given moment. At each moment you can either travel to another galaxy through one of wormholes that exist at this moment or you can simply wait for one galaxy day to see which wormholes will lead from your current position at the next day.

Your task is to find the expected value of time needed to travel from galaxy 1 to galaxy n, if you act in the optimal way. It is guaranteed that this expected value exists.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of galaxies within reach.

Then follows a matrix of n rows and n columns. Each element pij represents the probability that there is a wormhole from galaxy i to galaxy j. All the probabilities are given in percents and are integers. It is guaranteed that all the elements on the main diagonal are equal to 100.

Output

Print a single real value — the expected value of the time needed to travel from galaxy 1 to galaxy n if one acts in an optimal way. Your answer will be considered correct if its absolute or relative error does not exceed 10 −6 10   − 6 .

Namely: let’s assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if |a−b|max(1,b)≤10−6 | a − b | max ( 1 , b ) ≤ 10 − 6 .

Examples

input

3
100 50 50
0 100 80
0 0 100

output

1.750000000000000

input

2
100 30
40 100

output

3.333333333333333

Note

In the second sample the wormhole from galaxy 1 to galaxy 2 appears every day with probability equal to 0.3. The expected value of days one needs to wait before this event occurs is 10.3=313 1 0.3 = 3 1 3 .

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被dalao秒切的题。
然而咱太蒻切不掉……

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题意:
n n 个点的完全图，每条边有出现概率 pij p i j ，求从 1 1 号点到达 n n 号点的最优期望距离。

思路:
考虑这个最优。
设 f(x) f ( x ) 为当前从 x x 走到 n n 所需的最优期望距离。

可以发现，对于每个点，决策显然是优先走 f(x) f ( x ) 更小的点。
同时有一个性质:玩家绝对不会走回头路，也就是绝对不会往 f(x) f ( x ) 更小的节点走。

于是设 ai a i 表示 f(x) f ( x ) 第 i i 小的节点编号，有转移式:

f(ai)=1+∑j=1i(f(j)∗paiaj∗∏k=1j−1(1−paiak)) f ( a i ) = 1 + ∑ j = 1 i ( f ( j ) ∗ p a i a j ∗ ∏ k = 1 j − 1 ( 1 − p a i a k ) )

易知 a1=n a 1 = n 。
每确定一个 ai a i ，就更新每个节点的 f(x) f ( x ) ，取 f(x) f ( x ) 最小的节点作为 ai+1 a i + 1 ，以此类推。
这样从小到大确定每个节点的 f(x) f ( x ) 即可~

#include
using namespace std;

inline int read()
{
    int x=0;char ch=getchar();
    while(ch<'0' || '9'while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
    return x;
}

typedef double db;
const int N=1e3+9;

int n,q[N];
bool vis[N];
db p[N][N],f[N],g[N],h[N];

int main()
{
    n=read();
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            p[i][j]=(db)read()/100.0;

    for(int i=1;i<=n;i++)
        h[i]=0,g[i]=1;

    q[1]=n;vis[n]=1;
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
            if(!vis[j])
            {
                h[j]+=f[q[i-1]]*p[j][q[i-1]]*g[j];
                g[j]*=1-p[j][q[i-1]];
                f[j]=(1+h[j])/(1-g[j]);
            }

        for(int j=1;j<=n;j++)
            if(!vis[j] && (!q[i] || f[j]1;
    }

    printf("%.7lf\n",f[1]);
    return 0;
}