Description

Alas! A set of D (1 <= D <= 15) diseases (numbered 1..D) is running through the farm. Farmer John would like to milk as many of his N (1 <= N <= 1,000) cows as possible. If the milked cows carry more than K (1 <= K <= D) different diseases among them, then the milk will be too contaminated and will have to be discarded in its entirety. Please help determine the largest number of cows FJ can milk without having to discard the milk.

Input

* Line 1: Three space-separated integers: N, D, and K * Lines 2..N+1: Line i+1 describes the diseases of cow i with a list of 1 or more space-separated integers. The first integer, d_i, is the count of cow i‘s diseases; the next d_i integers enumerate the actual diseases. Of course, the list is empty if d_i is 0. 有N頭牛,它們可能患有D種病,現在從這些牛中選出若幹頭來,但選出來的牛患病的集合中不過超過K種病.

Output

* Line 1: M, the maximum number of cows which can be milked.

Sample Input

Sample Output

#include <cstdio>
#include <cstring>
char buf[10000000], *ptr = buf - 1;
inline int readint(){
    int f = 1, n = 0;
    char ch = *++ptr;
    while(ch < ‘0‘ || ch > ‘9‘){
        if(ch == ‘-‘) f = -1;
        ch = *++ptr;
    }
    while(ch <= ‘9‘ && ch >= ‘0‘){
        n = (n << 1) + (n << 3) + ch - ‘0‘;
        ch = *++ptr; 
    }
    return f * n;
}
const int maxn = 1000 + 10;
int s[maxn] = {0}, t[32768];
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    int N, D, K, Max;
    N = readint();
    D = readint();
    K = readint();
    Max = (1 << D) - 1;
    for(int x, i = 1; i <= N; i++){
        x = readint();
        while(x--) s[i] |= 1 << readint() - 1;
    }
    int ans = 0;
    t[0] = 0;
    for(int sum, i = 1; i <= Max; i++){
        t[i] = t[i ^ (i & -i)] + 1;
        if(t[i] > K) continue;
        sum = 0;
        for(int j = 1; j <= N; j++)
            if((i & s[j]) == s[j]) sum++;
        if(sum > ans) ans = sum;
    }
    printf("%d\n", ans);
    return 0;
}

[BZOJ1688][Usaco2005 Open]Disease Manangement 疾病管理