UVA - 489 Hangman Judge

Hangman Judge

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Submit Status

Description

In ``Hangman Judge,'' you are to write a program that judges a series of Hangman games. For each game, the answer to the puzzle is given as well as the guesses. Rules are the same as the classic game of hangman, and are given as follows:

1. The contestant tries to solve to puzzle by guessing one letter at a time.
2. Every time a guess is correct, all the characters in the word that match the guess will be ``turned over.'' For example, if your guess is ``o'' and the word is ``book'', then both ``o''s in the solution will be counted as ``solved.''
3. Every time a wrong guess is made, a stroke will be added to the drawing of a hangman, which needs 7 strokes to complete. Each unique wrong guess only counts against the contestant once.

      ______   
      |  |     
      |  O     
      | /|\    
      |  |     
      | / \    
    __|_       
    |   |______
    |_________|

4. If the drawing of the hangman is completed before the contestant has successfully guessed all the characters of the word, the contestant loses.
5. If the contestant has guessed all the characters of the word before the drawing is complete, the contestant wins the game.
6. If the contestant does not guess enough letters to either win or lose, the contestant chickens out.

Your task as the ``Hangman Judge'' is to determine, for each game, whether the contestant wins, loses, or fails to finish a game.

Input

Your program will be given a series of inputs regarding the status of a game. All input will be in lower case. The first line of each section will contain a number to indicate which round of the game is being played; the next line will be the solution to the puzzle; the last line is a sequence of the guesses made by the contestant. A round number of -1 would indicate the end of all games (and input).

Output

The output of your program is to indicate which round of the game the contestant is currently playing as well as the result of the game. There are three possible results:

You win.
You lose.
You chickened out.

Sample Input

1
cheese
chese
2
cheese
abcdefg
3
cheese
abcdefgij
-1

Sample Output

Round 1
You win.
Round 2
You chickened out.
Round 3
You lose.




分析：
题意：
         给定两个字符串，第一个串是用来匹配的，从第二个串的第一个字符开始匹配，如果第二个串中的字符在第一个串出现，则表示猜中了，第一个串中的相同的那个字符都算被猜中；如果没有出现则表示猜错，同样的猜错只算一次。在整个匹配的过程中，如果在还没猜错7次之前，第一个串中所有的字符都被猜完了， 则输出“You win.”，如果你还没全部猜完的时候就已经猜错7次，则输出“You lose.”。如果整个匹配过程结束后，你没赢也没输，则输出“You chickened out.”。
需要注意几点:
     1.若正确答案中某个字母出现了多次,那么只要猜中那个字母一次,就算答案中那个字母全被猜中.
     2.猜错的字母如果重复,则只算错一次.
     若7条命都没了,就算失败,全部猜中则算成功.如果没全猜中但是还有剩余生命,则算chikened out.
解法：
        先记录第一个串所有唯一的字符出现的次数，并保存在map中。然后扫描第二个串，实时记录猜对和猜错的情况，记录的时候不管是猜错还是猜对都要进行查重。每次匹配完一个字符就判断一下，是否是赢还是输。如果整个匹配过程下来都没输出赢或输。则输出“You chickened out.”。


代码：
#include <iostream>
#include <set>
#include <map>
#include <string.h>
#include <stdio.h>
using namespace std;


int main()
{
	int cas;
	int total;
	int r_cnt;
	int e_cnt;
	char s[1024];
	char g[1024];
	///统计每个字符出现的次数
	map<char, int> m;
	///猜错查重集合
	set<char> error;
	///猜对查重集合
	set<char> right;
	
	while (scanf("%d\n", &cas) != EOF) 
	{
		if (cas == -1) break;
		///每次的时候都清空容器
		m.clear();
		error.clear();
		right.clear();
		gets(s);
		gets(g);
		
		total = strlen(s);
		char * p = s;
		
		///统计每个字符出现的次数
		while (*p) 
		{
			if (m.find(*p) != m.end()) 
			{
				m[*p]++;
			}
			else 
			{
				m[*p] = 1;
			}
			p++;
		}
		
		r_cnt = 0;
		e_cnt = 0;
		
		p = g;
		int flag = 0;
		while (*p) 
		{
			///猜对
			if (m.find(*p) != m.end()) 
			{
				///之前没猜对过
				if (right.find(*p) == right.end()) 
				{
					r_cnt += m[*p];
					right.insert(*p);
				}
			}
			else ///猜错
			{
				///之前没猜错过
				if (error.find(*p) == error.end()) 
				{
					e_cnt++;
					error.insert(*p);
				}
			}
			
			if (r_cnt == total) 
			{
				flag = 1;
				break;
			}
			
			if (e_cnt >= 7) 
			{
				flag = 2;
				break;
			}
			p++;
		}
		
		cout << "Round " << cas << endl;
		
		switch(flag) 
		{
		case 0:
			cout << "You chickened out." << endl;
			break;
		case 1:
			cout << "You win." << endl;
			break;
		case 2:
			cout << "You lose." << endl;
			break;
		}
	}
	return 0;
}