二分贪心24

Description

The whole family was excited by the news. Everyone knew grandpa had been an extremely good bridge player for decades, but when it was announced he would be in the Guinness Book of World Records as the most successful bridge player ever, whow, that was astonishing! 

The International Bridge Association (IBA) has maintained, for several years, a weekly ranking of the best players in the world. Considering that each appearance in a weekly ranking constitutes a point for the player, grandpa was nominated the best player ever
because he got the highest number of points. 

Having many friends who were also competing against him, grandpa is extremely curious to know which player(s) took the second place. Since the IBA rankings are now available in the internet he turned to you for help. He needs a program which, when given a list
of weekly rankings, finds out which player(s) got the second place according to the number of points.

Input

The input contains several test cases. Players are identified by integers from 1 to 10000. The first line of a test case contains two integers N and M indicating respectively the number of rankings available (2 <= N <= 500) and the number of players in each
ranking (2 <= M <= 500). Each of the next N lines contains the description of one weekly ranking. Each description is composed by a sequence of M integers, separated by a blank space, identifying the players who figured in that weekly ranking. You can assume
that: 
in each test case there is exactly one best player and at least one second best player, 

each weekly ranking consists of M distinct player identifiers.

The end of input is indicated by N = M = 0.

Output

For each test case in the input your program must produce one line of output, containing the identification number of the player who is second best in number of appearances in t
4000
he rankings. If there is a tie for second best, print the identification numbers
of all second best players in increasing order. Each identification number produced must be followed by a blank space.

Sample Input

4 5
20 33 25 32 99
32 86 99 25 10
20 99 10 33 86
19 33 74 99 32
3 6
2 34 67 36 79 93
100 38 21 76 91 85
32 23 85 31 88 1
0 0

Sample Output

32 33

1 2 21 23 31 32 34 36 38 67 76 79 88 91 93 100
题目大意：给一个n*m的矩阵，输出出现次数第二多的全部数据，并且按照从小到大顺序
#include <iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
struct dalao
{
int t;
int ans;
}num[10000+10];
bool cmp(dalao a,dalao b)
{
return a.ans>b.ans;
}
bool cmp2(dalao a,dalao b)
{
return a.t<b.t;
}
int main()
{
int n,m,i,j;
while(scanf("%d %d",&n,&m))
{
if(m==0&&n==0)break;
memset(num,0,sizeof(num));
for(i=0;i<m*n;i++)
{
scanf("%d",&j);
num[j].t=j;
num[j].ans++;
}
int k=1;
sort(num,num+10000,cmp);
for(i=2;i<m*n;i++)
if(num[i].ans!=num[<
b215
/span>1].ans)break;
else k++;
sort(num+1,num+k+1,cmp2);
printf("%d",num[1].t);
for(i=2;i<=k;i++)
printf(" %d",num[i].t);
printf("\n");
}
return 0;
}