XVII Open Cup named after E.V. Pankratiev. XXI Ural Championship

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N = 55555;
int n, m, i, j, k, x, ans[N], mx[N], b[N];
vector<int>a[N];
bool v[(1 << 26) + 5];
inline int get() //!!!!二进制压缩  把每个人的口味压缩成1<<26内的数
{
        static char s[1000];
        scanf("%s", s);
        int t = 0, len = strlen(s);
        for (int i = 0; i < len; i++)
        {
                t |= 1 << (s[i] - ‘a‘); //!!!!!!转换成二进制
        }
        return t;
}
inline void make(int x, int y)
{
        for (int i = x; i; i = (i - 1)&x) //!!!!枚举每位1存在或不存在的情况
        {
                v[i] = y;
        }
}
int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);
        scanf("%d", &n);
        for (i = 1; i <= n; i++)
        {
                scanf("%d", &k);
                while (k--)
                {
                        x = get();
                        a[i].push_back(x);
                }
                ans[i] = 1;
        }
        scanf("%d", &m);
        for (i = 1; i <= m; i++)
        {
                scanf("%d", &k);
                for (j = 1; j <= k; j++)
                {
                        b[j] = get();
                }
                for (j = 1; j <= k; j++)
                {
                        make(b[j], 1);  //!!!找出每种情况
                }
                for (j = 1; j <= n; j++)
                {
                        int t = 0;
                        for (x = 0; x < a[j].size(); x++)
                                if (v[a[j][x]])
                                {
                                        t++;
                                }
                        if (t > mx[j])
                        {
                                mx[j] = t, ans[j] = i;//!!!!找出每个group最适应的店
                        }
                }
                for (j = 1; j <= k; j++)
                {
                        make(b[j], 0);  //!!!相当于memset 把先前变为1的变回0
                }
        }
        for (i = 1; i <= n; i++)
        {
                printf("%d\n", ans[i]);
        }
}