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2013
11-09

POJ 1323 Game Prediction [解题报告] Java

Game Prediction

问题描述 :

Suppose there are M people, including you, playing a special card game. At the beginning, each player receives N cards. The pip of a card is a positive integer which is at most N*M. And there are no two cards with the same pip. During a round, each player chooses one card to compare with others. The player whose card with the biggest pip wins the round, and then the next round begins. After N rounds, when all the cards of each player have been chosen, the player who has won the most rounds is the winner of the game.

Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game.

输入:

The input consists of several test cases. The first line of each case contains two integers m (2?20) and n (1?50), representing the number of players and the number of cards each player receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases.

The input is terminated by a line with two zeros.

输出:

For each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game.

样例输入:

2 5
1 7 2 10 9

6 11
62 63 54 66 65 61 57 56 50 53 48

0 0

样例输出:

Case 1: 2
Case 2: 4

解题代码:

//* @author 洪晓鹏<[email protected]>
import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int time = 1; while(true) { int m = in.nextInt(); int n = in.nextInt(); if(m==0 && n==0) { break; } int[] cards = new int[n]; boolean[] isUsed = new boolean[n*m+1]; for(int i = 0; i< n; i++) { cards[i] = in.nextInt(); isUsed[ cards[i]]=true; } Arrays.sort(cards); int result = n; for(int i = n-1; i>=0; i--) { int p=cards[i]; if(p==n*m) continue; while(p < n*m) { if(isUsed[p+1] == false) { isUsed[p+1] = true; result--; break; } else p++; } } System.out.println("Case "+time+": "+result); time++; } } }

  1. 学算法中的数据结构学到一定程度会乐此不疲的,比如其中的2-3树,类似的红黑树,我甚至可以自己写个逻辑文件系统结构来。

  2. A猴子认识的所有猴子和B猴子认识的所有猴子都能认识,这句话用《爱屋及乌》描述比较容易理解……

  3. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。