[HDU5955]Guessing the Dice Roll

Problem Description

There are N players playing a guessing game. Each player guesses a sequence consists of {1,2,3,4,5,6} with length L, then a dice will be rolled again and again and the roll out sequence will be recorded. The player whose guessing sequence first matches the last L rolls of the dice wins the game.

 

Input

The first line is the number of test cases. For each test case, the first line contains 2 integers N (1 ≤ N ≤ 10) and L (1 ≤ L ≤ 10). Each of the following N lines contains a guessing sequence with length L. It is guaranteed that the guessing sequences are consist of {1,2,3,4,5,6} and all the guessing sequences are distinct.

 

Output

For each test case, output a line containing the winning probability of each player with the precision of 6 digits.

 

Sample Input

3 5 1 1 2 3 4 5 6 2 1 1 2 1 3 1 4 1 5 1 6 1 4 3 1 2 3 2 3 4 3 4 5 4 5 6

 

Sample Output

0.200000 0.200000 0.200000 0.200000 0.200000 0.027778 0.194444 0.194444 0.194444 0.194444 0.194444 0.285337 0.237781 0.237781 0.239102

 

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对所有串建AC自动机，然后按照Trie图建立递推关系，转化成矩阵，用高斯消元解决带环的转移。

写起来略恶心，反正我是不想调了。

 

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
#define reg register
inline int read() {
    int res = 0;char ch=getchar();bool fu=0;
    while(!isdigit(ch))fu|=(ch=='-'),ch=getchar();
    while(isdigit(ch)) res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
    return fu?-res:res;
}

int T;
int n, L;
int nxt[105][7], fail[105], danger[105], tot, who[105];
inline void ins(int *s, int id) 
{
    int now = 0;
    for (reg int i = 1 ; i <= L ; i ++)
        now = nxt[now][s[i]] ? nxt[now][s[i]] : nxt[now][s[i]] = ++tot;
    danger[now] = 1;
    who[now] = id;
}

inline void Build()
{
    queue <int> q;
    for (reg int i = 1 ; i <= 6 ;i ++) if (nxt[0][i]) q.push(nxt[0][i]);
    while(!q.empty())
    {
        int x = q.front();q.pop();
        for (reg int i = 1 ; i <= 6 ; i ++)
            if (nxt[x][i]) fail[nxt[x][i]] = nxt[fail[x]][i], q.push(nxt[x][i]);
            else nxt[x][i] = nxt[fail[x]][i];
        danger[x] |= danger[fail[x]];
    }
}

long double a[105][105], ans[15];

inline void Gauss()
{
    for (reg int i = 0 ; i <= tot ; i ++)
    {
        int k = i;
        for (reg int j = i ; j <= tot ; j ++)
            if (fabs(a[j][i]) > fabs(a[k][i])) k = j;
        if (k != i) swap(a[k], a[i]);
        for (reg int j = 0 ; j <= tot ; j ++)
        {
            if (i == j) continue;
            long double r = a[j][i] / a[i][i];
            for (reg int k = i ; k <= tot + 1 ; k ++)
                a[j][k] -= a[i][k] * r;
        }
    }
}

int main()
{
    T = read();
    while(T--)
    {
        memset(nxt, 0, sizeof nxt);
        memset(fail, 0, sizeof fail);
        memset(danger, 0, sizeof danger);
        tot = 0;
        n = read(), L = read();
        int tmp[13];
        for (reg int i = 1 ; i <= n ; i ++)
        {
            for (reg int j = 1 ; j <= L ; j ++) tmp[j] = read();
            ins(tmp, i);
        }
        Build();
        memset(a, 0, sizeof a);
        for (reg int i = 0 ; i <= tot ; i ++)
        {
            a[i][i] -= 1.0;
            if (danger[i]) continue;
            for (reg int j = 1 ; j <= 6 ; j ++)
                a[nxt[i][j]][i] += 1.0 / 6.0;
        }
        a[0][tot + 1] = -1.0;
//        for(int i=0;i<=tot;i++,puts(""))
//            for(int j=0;j<=tot+1;j++) printf("%.2Lf ",a[i][j]);
        Gauss();
        for (reg int i = 0 ; i <= tot ; i ++)
            if (danger[i]) ans[who[i]] = a[i][tot + 1] / a[i][i];
        for (reg int i = 1 ; i < n ; i ++) printf("%.6Lf ", ans[i]);
        printf("%.6Lf
", ans[n]);
    }
    return 0;
}