(贪心)AtCoder Grand Contest 018 B : Sports Festival

Problem Statement

Takahashi is hosting an sports meet. There are N people who will participate. These people are conveniently numbered 1 through N. Also, there are M options of sports for this event. These sports are numbered 1 through M. Among these options, Takahashi will select one or more sports (possibly all) to be played in the event.

Takahashi knows that Person i's j-th favorite sport is Sport Aij. Each person will only participate in his/her most favorite sport among the ones that are actually played in the event, and will not participate in the other sports.

Takahashi is worried that one of the sports will attract too many people. Therefore, he would like to carefully select sports to be played so that the number of the participants in the sport with the largest number of participants is minimized. Find the minimum possible number of the participants in the sport with the largest number of participants.

Constraints

  • 1≤N≤300
  • 1≤M≤300
  • Ai1 , Ai2 ,  , AiM is a permutation of the integers from 1 to M.

Input

Input is given from Standard Input in the following format:

N M
A11 A12  A1M
A21 A22  A2M
:
AN1 AN2  ANM

Output

Print the minimum possible number of the participants in the sport with the largest number of participants.


Sample Input 1

Copy
4 5
5 1 3 4 2
2 5 3 1 4
2 3 1 4 5
2 5 4 3 1

Sample Output 1

Copy
2

Assume that Sports 13 and 4 are selected to be played. In this case, Person 1 will participate in Sport 1, Person 2 in Sport 3, Person 3 in Sport 3 and Person 4 in Sport 4. Here, the sport with the largest number of participants is Sport 3, with two participants. There is no way to reduce the number of participants in the sport with the largest number of participants to 1. Therefore, the answer is 2.


Sample Input 2

Copy
3 3
2 1 3
2 1 3
2 1 3

Sample Output 2

Copy
3

Since all the people have the same taste in sports, there will be a sport with three participants, no matter what sports are selected. Therefore, the answer is 3.

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <queue>
 8 #include <set>
 9 #include <map>
10 #include <list>
11 #include <vector>
12 #include <stack>
13 #define mp make_pair
14 #define MIN(a,b) (a>b?b:a)
15 //#define MAX(a,b) (a>b?a:b)
16 typedef long long ll;
17 typedef unsigned long long ull;
18 const int MAX=2e5+5;
19 const int INF=1e8+5;
20 using namespace std;
21 //const int MOD=1e9+7;
22 typedef pair<ll,int> pii;
23 const double eps=0.00000001;
24 int n,m;
25 queue<int> que[305];
26 int tem;
27 bool vi[305];
28 int num[305];
29 int cnt,an,who;
30 int main()
31 {
32     scanf("%d%d",&n,&m);
33     for(int i=1;i<=n;i++)
34     {
35         for(int j=1;j<=m;j++)
36         {
37             scanf("%d",&tem);
38             que[i].push(tem);
39         }
40     }
41     for(int i=1;i<=m;i++)
42     {
43         memset(num,0,sizeof(num));
44         cnt=0;
45         who=que[1].front();
46         for(int j=1;j<=n;j++)
47         {
48             int pre=que[j].front();
49             while(vi[pre])
50             {
51                 que[j].pop();
52                 pre=que[j].front();
53             }
54             ++num[pre];
55             if(num[pre]>cnt)
56             {
57                  cnt=num[pre];
58                  who=pre;
59             }
60         }
61         vi[who]=true;
62         if(i!=1)
63             an=min(an,cnt);
64         else
65             an=cnt;
66     }
67     printf("%d
",an);
68     return 0;
69 }
原文地址:https://www.cnblogs.com/quintessence/p/7226767.html