hdu5955 Guessing the Dice Roll【AC自动机】【高斯消元】【概率】

含高斯消元模板

2016沈阳区域赛http://acm.hdu.edu.cn/showproblem.php?pid=5955

Guessing the Dice Roll

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1632    Accepted Submission(s): 480

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

 

Source

2016ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

Recommend

jiangzijing2015

 

 

 

题意：

有n个人猜掷骰子的序列。给定序列长度是l。只要n个人中的一个序列出现了，这个人就赢了游戏结束。问他们获胜的概率是多少。

思路：

没有想到是AC自动机。但是其实他本质就是在一个自动机的不同状态之间转转转，看最后转到哪个状态上去。

对这几个序列建立了AC自动机之后，就可以列出他们互相之间转移的概率方程了。然后解方程就可以得到每个人获胜的概率。

矩阵中的第(j)行表示的就是关于(j)这个状态的方程。(a[j][i])表示由状态(i)转移到状态(j)的概率。

(a[i][i])本身应该放在方程的右边，所以他是(-1)

根节点是比较特殊的，我们需要再设置一个虚拟节点，虚拟节点到根节点的概率是1，他也应该作为根节点初始时的概率放在右边。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<set>
using namespace std;
typedef long long LL;
#define N 100010
#define pi 3.1415926535

int n, l, t;
const int maxn = 105;
struct trie{
    int son[7];
    int ed;
    int fail;
}AC[maxn];
int tot = 0;
int fp[11];

void build(int s[], int id)
{
    int now = 0;
    for(int i = 0; i < l; i++){
        if(AC[now].son[s[i]] == 0){
            AC[now].son[s[i]] = ++tot;
        }
        now = AC[now].son[s[i]];
    }
    AC[now].ed = id;
    fp[id] = now;
}

void get_fail()
{
    queue<int>que;
    for(int i = 1; i <= 6; i++){
        if(AC[0].son[i] != 0){
            AC[AC[0].son[i]].fail = 0;
            que.push(AC[0].son[i]);
        }
    }

    while(!que.empty()){
        int u = que.front();
        que.pop();
        for(int i = 1; i <= 6; i++){
            if(AC[u].son[i] != 0){
                AC[AC[u].son[i]].fail = AC[AC[u].fail].son[i];
                que.push(AC[u].son[i]);
            }
            else{
                AC[u].son[i] = AC[AC[u].fail].son[i];
            }
        }
    }
}

double a[maxn][maxn], x[maxn];
const double eps = 1e-10;
int equ, var;
void gauss()
{
    equ = var = tot + 1;
    int i,j,k,col,max_r;
    for(k=0,col=0;k<equ&&col<var;k++,col++)
    {
        max_r=k;
        for(i=k+1;i<equ;i++)
        {
            if(fabs(a[i][col] )>fabs(a[max_r][col] ) ) max_r=i;
        }
        if(fabs(a[max_r][col])<eps )  return;
        if(k!=max_r)
        {
            for(j=col;j<=var;j++)  swap(a[k][j],a[max_r][j]  );
        }
        for(j=col+1;j<=var;j++)  a[k][j]/=a[k][col];

        a[k][col]=1;

        for(i=0;i<equ;i++) if(i!=k)
        {
            for(j=col+1;j<=var;j++) a[i][j]-=a[k][j]*a[i][col];

            a[i][col]=0;
        }
    }
    for(i=0;i<equ;i++)  x[i]=a[i][var];
    return;
}

int main()
{
    scanf("%d", &t);
    while(t--){
        for(int i = 0; i <= tot; i++){
            AC[i].fail = 0;
            AC[i].ed = 0;
            for(int j = 0; j < 7; j++){
                AC[i].son[j] = 0;
            }
        }
        tot = 0;

        scanf("%d%d", &n, &l);
        for(int i = 1; i <= n; i++){
            int tmp[11];
            for(int j = 0; j < l; j++){
                scanf("%d", &tmp[j]);
            }
            build(tmp, i);
        }
        get_fail();

        memset(a, 0, sizeof(a));
        memset(x, 0, sizeof(x));
        for(int i = 0; i <= tot; i++)a[i][i] = -1.0;
        for(int i = 0; i <= tot; i++){
            if(AC[i].ed == 0){
                for(int j = 1; j <= 6; j++){
                    int to = AC[i].son[j];
                    a[to][i] += 1.0 / 6;
                }
            }

        }

        a[0][tot + 1] = -1.0;//虚拟节点
        gauss();
        for(int i = 1; i <= n; i++){
            printf("%.6f", x[fp[i]]);
            if(i == n){
                printf("
");
            }
            else{
                printf(" ");
            }
        }

        //cout<<"yes"<<endl;
    }

    return 0;
}