CodeForces

Description

Lately, a national version of a bingo game has become very popular in Berland. There are n players playing the game, each player has a card with numbers. The numbers on each card are distinct, but distinct cards can have equal numbers. The card of the i-th player contains mi numbers.

During the game the host takes numbered balls one by one from a bag. He reads the number aloud in a high and clear voice and then puts the ball away. All participants cross out the number if it occurs on their cards. The person who crosses out all numbers from his card first, wins. If multiple people cross out all numbers from their cards at the same time, there are no winners in the game. At the beginning of the game the bag contains 100 balls numbered 1 through 100, the numbers of all balls are distinct.

You are given the cards for each player. Write a program that determines whether a player can win the game at the most favorable for him scenario or not.

Input

The first line of the input contains integer n (1 ≤ n ≤ 100) — the number of the players. Then follow n lines, each line describes a player’s card. The line that describes a card starts from integer mi (1 ≤ mi ≤ 100) that shows how many numbers the i-th player’s card has. Then follows a sequence of integers ai, 1, ai, 2, …, ai, mi (1 ≤ ai, k ≤ 100) — the numbers on the i-th player’s card. The numbers in the lines are separated by single spaces.

It is guaranteed that all the numbers on each card are distinct.

Output

Print n lines, the i-th line must contain word “YES” (without the quotes), if the i-th player can win, and “NO” (without the quotes) otherwise.

Sample Input

3
1 1
3 2 4 1
2 10 11

2
1 1
1 1

Sample Output

YES
NO
YES

NO
NO

Hint

题意

一个人报1到100的数 n个人参加游戏 每个人手上的卡片有一些数字 只要一个人的数字先被报完 他就赢了 如果有人和他同时获胜 那他就输了 对于每个人判断的时候都是最佳情况（即先报的数都是他有的数）

题解：

一个串没有的 另一个串也没有 就说明这另一个串是这个串的子串

AC代码

#include <cstdio>#include <queue>#include <cstring>#include <algorithm>using namespace std;int a[111][111];int main(){    int n;    scanf("%d",&n);    int m;    int x;    for (int i = 0; i < n; ++i){        scanf("%d",&m);        for (int j = 0; j < m; ++j){            scanf("%d",&x);            a[i][x] = 1;        }    }    for (int i = 0; i < n; ++i){        int flag = 1;        for (int j = 0 ; j < n; ++j){            if (i!=j){                int t = 1;                for (int k = 1; k <= 100; ++k){                    if (!a[i][k]&&a[j][k]) {t = 0; break;}                }                if (t) {flag=0;break;}            }        }        if (flag) printf("YES\n");else printf("NO\n");    }    return 0;}