POJ 2092 Grandpa is Famous【水---找出现第二多的数】
链接:
Time Limit: 2000MS | Memory Limit: 30000K | |
Total Submissions: 7210 | Accepted: 3650 |
Description
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
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
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
Source
Code:
/**************************************************************
D Accepted 284 KB 204 ms C++ 1604 B
题意:第一行给你 N 和 M
剩下的 N 行 M 列, 给你 N*M 个数
找出所有出现次数第二多的数, 并且按照从小到大顺序输出。
**************************************************************/
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 10000+10; int a[maxn];
struct Node{
int num;
int index;
}node[maxn]; bool cmp(Node a, Node b)
{
if(a.num == b.num) return a.index < b.index;
else return a.num > b.num;
} int main()
{
int n,m;
while(scanf("%d%d", &n,&m) != 0)
{
if(n == 0 && m == 0) break;
int x;
int Max = 0; for(int i = 0; i < maxn; i++)
{
a[i] = 0;
node[i].num = 0;
node[i].index = 0;
} int N = n*m;
for(int i = 0; i < N; i++)
{
scanf("%d", &x);
a[x]++;
Max = max(Max, x); }
N = Max+1;
//printf("Max = %d\n", Max);
int j = 0;
for(int i = 1; i < N; i++)
{
if(a[i] != 0)
{
node[j].num = a[i];
node[j].index = i;
j++;
}
}
sort(node,node+j,cmp);
N = j; int s = node[0].num; //printf("s1 = %d\n", s);
int index = 0;
for(int i = 0; i < N; i++)
{
if(node[i].num < s)
{
s = node[i].num;
index = i;
break;
}
}
//printf("s2 = %d\n", s);
for(int i = index; i < N; i++)
{
if(node[i].num == s)
{
printf("%d ", node[i].index);
}
else if(node[i].num > s) break;
}
printf("\n");
}
return 0;
}
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