1121. Damn Single (25)

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原文链接：http://www.cnblogs.com/childwang/p/8280269.html

本文解析PAT算法竞赛中的一道题目“单身狗”，通过使用C++实现，介绍如何利用set数据结构来高效解决该问题。文章详细阐述了输入输出规格，并给出样例及解答思路。

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1121. Damn Single (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

"Damn Single (单身狗)" is the Chinese nickname for someone who is being single. You are supposed to find those who are alone in a big party, so they can be taken care of.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=50000), the total number of couples. Then N lines of the couples follow, each gives a couple of ID's which are 5-digit numbers (i.e. from 00000 to 99999). After the list of couples, there is a positive integer M (<=10000) followed by M ID's of the party guests. The numbers are separated by spaces. It is guaranteed that nobody is having bigamous marriage (重婚) or dangling with more than one companion.

Output Specification:

First print in a line the total number of lonely guests. Then in the next line, print their ID's in increasing order. The numbers must be separated by exactly 1 space, and there must be no extra space at the end of the line.

Sample Input:

3
11111 22222
33333 44444
55555 66666
7
55555 44444 10000 88888 22222 11111 23333

Sample Output:

5
10000 23333 44444 55555 88888

题意：输入n对夫妻的名单，再输入m个客人列表，判断今晚有多少人单身（已婚的，但是对方没来算单身）

分析：圣诞节做single dog的题真带感(雾

先记录下夫妻的名单，输入客人时，若单身，则加入set中；若已婚并对象未出现，则加入set中；若已婚并对象已出现，则利用set的find，删除他们。

set的find还是很快的，20000个也不过14次查找。虽然暴力了一点，但实现很方便。

另外需要注意的是，因为00000也是客人的编号，所以夫妻名单要初始化成-1，用0的话会产生歧义（一夫多多多多妻）

代码：

#include<cstdio>
#include<set>
#include<cstring>
using namespace std;
int couple[100050];
set<int> single;
int main()
{
	int n,m,a,b;
	scanf("%d",&n);
	memset(couple,-1,sizeof(int)*100050);
	for(int i=0;i<n;i++)
	{
		scanf("%d %d",&a,&b);
		couple[a]=b;
		couple[b]=a;
	}
	scanf("%d",&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d",&a);
		if(couple[a]==-1) single.insert(a);
		else{
			if(single.find(couple[a])!=single.end())
				single.erase(single.find(couple[a]));
			else single.insert(a);
		}
	}
	printf("%d\n",single.size());
	set<int>::iterator it;
	for(it=single.begin();it!=single.end();it++)
		it==single.begin()?printf("%05d",*it):printf(" %05d",*it);
}

结尾：感觉pat里对set考察挺多的

转载于:https://www.cnblogs.com/childwang/p/8280269.html