首页 > 数据结构 > Hash表 > POJ 2092 Grandpa is Famous [解题报告] Java
2013
11-10

POJ 2092 Grandpa is Famous [解题报告] Java

Grandpa is Famous

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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

解题代码:

//* @author: [email protected]
import java.io.*;
import java.util.*;
import java.util.Map.Entry;
public class Main
{
 public static void main(String[] args) throws NumberFormatException, IOException
 {
  InputStreamReader is=new InputStreamReader(System.in);
  BufferedReader in=new BufferedReader(is);
  HashMap< Integer,Integer> ts=new HashMap< Integer,Integer>();
  while(true)
  {
   String[] ss=in.readLine().split(" ");
   int x=Integer.parseInt(ss[0]);
   int y=Integer.parseInt(ss[1]);
   if(x==0&&y==0)break;
   for(int i=0;i< x;i++)
   {
    ss=in.readLine().split(" ");
    for(int j=0;j< y;j++)
    {
      int u=Integer.parseInt(ss[j]);
      ts.put(u, ts.containsKey(u)?ts.get(u)+1:1);
    }
    }
    List< Map.Entry< Integer, Integer>> ww=
           new ArrayList< Map.Entry< Integer, Integer>>(ts.entrySet());
    Collections.sort(ww,new Comparator< Map.Entry< Integer, Integer>>(){

    @Override
	public int compare(Entry< Integer, Integer> arg0,
	Entry< Integer, Integer> arg1) {
	  int r1=arg1.getValue();
	  int r0=arg0.getValue();
	  if(r1!=r0) return r1-r0;
	  else return arg0.getKey()-arg1.getKey();
        }
     });
	int k=ww.get(1).getValue();
	System.out.print(ww.get(1).getKey());
	for(int i=2;i< ww.size();i++)
	{
	  if(ww.get(i).getValue()==k) System.out.print(" "+ww.get(i).getKey());
	  else break;
	}
	System.out.println();
	ts.clear();
  }	
 }
}

  1. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])

  2. #include <stdio.h>
    int main()
    {
    int n,p,t[100]={1};
    for(int i=1;i<100;i++)
    t =i;
    while(scanf("%d",&n)&&n!=0){
    if(n==1)
    printf("Printing order for 1 pages:nSheet 1, front: Blank, 1n");
    else {
    if(n%4) p=n/4+1;
    else p=n/4;
    int q=4*p;
    printf("Printing order for %d pages:n",n);
    for(int i=0;i<p;i++){
    printf("Sheet %d, front: ",i+1);
    if(q>n) {printf("Blank, %dn",t[2*i+1]);}
    else {printf("%d, %dn",q,t[2*i+1]);}
    q–;//打印表前
    printf("Sheet %d, back : ",i+1);
    if(q>n) {printf("%d, Blankn",t[2*i+2]);}
    else {printf("%d, %dn",t[2*i+2],q);}
    q–;//打印表后
    }
    }
    }
    return 0;
    }