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| Time Limit: 1000MS |  | Memory Limit: 30000K | 
| Total Submissions: 5158 |  | Accepted: 2844 | 
 
Description
The ICPC committee would like to have its meeting as soon as possible to address every little issue of the next contest. However, members of the committee are so busy maniacally developing (possibly useless) programs that it is very difficult to arrange their schedules for the meeting. So, in order to settle the meeting date, the chairperson requested every member to send back a list of convenient dates by E-mail. Your mission is to help the chairperson, who is now dedicated to other issues of the contest, by writing a program that chooses the best date from the submitted lists. Your program should find the date convenient for the most members. If there is more than one such day, the earliest is the best.
Input
The input has multiple data sets, each starting with a line containing the number of committee members and the quorum of the meeting. 
N Q
Here, N, meaning the size of the committee, and Q meaning the 
quorum, are positive integers. N is less than 50, and, of course, Q is 
less than or equal to N. 
N lines follow, each describing convenient dates for a committee member in the following format. 
M Date1 Date2 ... DateM 
Here, M means the number of convenient dates for the member, which 
is an integer greater than or equal to zero. The remaining items in the 
line are his/her dates of convenience, which are positive integers less 
than 100, that is, 1 means tomorrow, 2 means the day after tomorrow, and
 so on. They are in ascending order without any repetition and separated
 by a space character. Lines have neither leading nor trailing spaces. 
A line containing two zeros indicates the end of the input. 
Output
For 
each data set, print a single line containing the date number convenient
 for the largest number of committee members. If there is more than one 
such date, print the earliest. However, if no dates are convenient for 
more than or equal to the quorum number of members, print 0 instead.
Sample Input
3 2
2 1 4
0
3 3 4 8
3 2
4 1 5 8 9
3 2 5 9
5 2 4 5 7 9
3 3
2 1 4
3 2 5 9
2 2 4
3 3
2 1 2
3 1 2 9
2 2 4
0 0
Sample Output
4
5
0
2
CODE:
#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 100 + 10
using namespace std;
int main(){
    int n, q, date[MAX_N];
    while(scanf("%d%d", &n, &q) != EOF){
        if(n == 0 && q == 0) break;
        int m, x, mdate = 0;
        memset(date, 0, sizeof(date));
        REP(i, 1, n){
            scanf("%d", &m);
            REP(i, 1, m){
                scanf("%d", &x), date[x] ++;
                if(x > mdate) mdate = x;
            }
        }
        int ans = 0;
        REP(i, 1, mdate){
            if(date[i] > date[ans]) ans = i;
        }
        if(date[ans] >= q) printf("%d\n", ans);
        else printf("0\n");
    }
    return 0;
}
 
 
POJ 2028 When Can We Meet?
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原文地址:http://www.cnblogs.com/ALXPCUN/p/4535478.html