Codeforces Round #444 (Div. 2) B. Cubes for Masha

B. Cubes for Masha

Problem Statement

    Absent-minded Masha got set of n cubes for her birthday.
    At each of 6 faces of each cube, there is exactly one digit from 0 to 9. Masha became interested what is the largest natural x such she can make using her new cubes all integers from 1 to x.
    To make a number Masha can rotate her cubes and put them in a row. After that, she looks at upper faces of cubes from left to right and reads the number.
    The number can’t contain leading zeros. It’s not required to use all cubes to build a number.
    Pay attention: Masha can’t make digit 6 from digit 9 and vice-versa using cube rotations.

Input

    In first line integer n is given (1 ≤ n ≤ 3) — the number of cubes, Masha got for her birthday.
    Each of next n lines contains 6 integers ai,j (0 ≤  ai,j  ≤ 9) — number on j-th face of i-th cube.

Output

    Print single integer — maximum number x such Masha can make any integers from 1 to x using her cubes or 0 if Masha can’t make even 1.

Examples

Example 1
    Input
        3
        0 1 2 3 4 5
        6 7 8 9 0 1
        2 3 4 5 6 7
    Output
        87
Example 2
    Input
        3
        0 1 3 5 6 8
        1 2 4 5 7 8
        2 3 4 6 7 9

Note

    In the first test case, Masha can build all numbers from 1 to 87, but she can’t make 88 because there are no two cubes with digit 8.

题意

    一个人有n个魔方，每个魔方的6个面上都有一个数字，定义一个数字可以被摆出为用该数字的数字位数个魔方并且经过重排列后可以从左往右看获得这个数。求最大的x使得1~x都可以被摆出。

思路

    因为最多只有3个魔方，所以即使每个魔方都有10面，那么能摆出的数字最大也只有999，所以我们可以很愉快地打暴力。对于每个数，我们暴力分解出每一位上的数字，然后dfs一遍是否有合法的解，直到找到第一个不能摆出的数。
    P.S. 一个魔方上的数不能在摆一个数的时候用多次。

Code

#pragma GCC optimize(3)
#include
using namespace std;
typedef long long ll;
bool Finish_read;
template<class T>inline void read(T &x){Finish_read=0;x=0;int f=1;char ch=getchar();while(!isdigit(ch)){if(ch=='-')f=-1;if(ch==EOF)return;ch=getchar();}while(isdigit(ch))x=x*10+ch-'0',ch=getchar();x*=f;Finish_read=1;}
template<class T>inline void print(T x){if(x/10!=0)print(x/10);putchar(x%10+'0');}
template<class T>inline void writeln(T x){if(x<0)putchar('-');x=abs(x);print(x);putchar('\n');}
template<class T>inline void write(T x){if(x<0)putchar('-');x=abs(x);print(x);}
/*================Header Template==============*/
int n,x,cnt[4][10],now[10],vis[4],pos;
inline bool dfs(int p) {
    if(p==pos+1)
        return 1;
    bool ans=0;
    for(int i=1;i<=n;i++) {
        if(!vis[i]&&cnt[i][now[p]]) {
            vis[i]=1;
            ans|=dfs(p+1);
            vis[i]=0;
        }
    }
    return ans;
}
int main() {
    read(n);
    for(int i=1;i<=n;i++) {
        for(int j=1;j<=6;j++) {
            read(x);
            cnt[i][x]++;
        }
    }
    for(int i=1;i<=1000;i++) {
        if(i==1000) {
            writeln(999);
            return 0;
        }
        int tmp=i;
        pos=0;
        while(tmp) {
            now[++pos]=tmp%10;
            tmp/=10;
        }
        if(!dfs(1)) {
            writeln(i-1);
            return 0;
        }
    }
}