Vjudge B - Grandpa is Famous

题目：

B - Grandpa is Famous
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 the 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.

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

Output：

32 33
1 2 21 23 31 32 34 36 38 67 76 79 88 91 93 100

题目大意：

输入一个n行m列的数组，只有一个出现次数最多的值，可能有多个出现次数第二多的值，找到数组中出现次数第二多的数并输出。输入n=0、m=0时结束。

实现代码及思想：

定义一个结构体数组，结构体里分别num和score两个变量，num代表选手的编号，score代表出现次数。然后对结构体数组p进行排序，因为只有一个出现次数最多的值，所以p[0].score是出现次数最多的值的次数，p[1].score是出现次数第二多的值的次数，找到第二多的次数，然后输出编号（num），再继续向后查找，如果次数等于p[1].score就输出编号（num）.

#include <iostream>
#include<algorithm>using namespace std;
const int MAXL = 10001;
int N,M;
struct player
{int num;int score;
}p[MAXL];bool cmp(player p1,player p2)
{if(p1.score!=p2.score)return p1.score>p2.score;//如果分数不同，分数高的在前elsereturn p1.num<p2.num;//如果分数相同，按照编号小的在前
}int main()
{  while(cin >> N >> M){if(N==0&&M==0)break;for(int i=0;i<MAXL;i++){p[i].score = 0;}int s;for(int i=0;i<N;i++){for(int j=0;j<M;j++){cin >> s;p[s].num =s;p[s].score++;  }}sort(p,p+MAXL,cmp);cout << p[1].num;for(int i=2;i<MAXL;i++){if(p[i].score==p[1].score){cout << " " << p[i].num;  }   }cout << endl;}return 0;
}