【PAT】1121. Damn Single (25)【哈希表】

题目描述

“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.

翻译：单身狗是中国对单身的人的绰号。你需要找出那些在一场大型派对上单身的人，这样他们可以就会被关照了。

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 50,000), 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 (≤ 10,000) 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.

翻译：每个输入文件包含一组测试数据。对于每组测试数据，第一行包括两个正整数N (≤ 50,000)，表示情侣的总人数。接下来给出N行情侣，每一行给出一对情侣的五位数字编号 (i.e. from 00000 to 99999)。在情侣列表后面，包含一个正整数M (≤ 10,000) ，跟着M个派对嘉宾的编号。数字之间用空格隔开。数据保证没有人有重婚或和多名女性拍拖的情况。

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.

翻译：输出一行单身嘉宾的数量。接下来，一行，按照升序输出它们的ID。数字之间必须用空格隔开，并且行末尾不能有多余空格。

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

解题思路

用哈希表记录每对情侣。然后对嘉宾数组进行排序，如果情侣双方都在则跳过。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#define INF 99999999
using namespace std;
int N,Hash[100000];
int M,Party[10010];
int v[50010];//记录couple对来的人
int main(){scanf("%d",&N);int a,b;for(int i=1;i<=N;i++){scanf("%d%d",&a,&b);Hash[a]=i;Hash[b]=i;}scanf("%d",&M);int ccount=M; for(int i=0;i<M;i++){scanf("%d",&Party[i]);if(Hash[Party[i]])v[Hash[Party[i]]]++;if(v[Hash[Party[i]]]>1)ccount-=2;}sort(Party,Party+M);printf("%d\n",ccount);int flag=0;for(int i=0;i<M;i++){if(v[Hash[Party[i]]]<=1){if(flag==0)printf("%05d",Party[i]),flag=1;else    printf(" %05d",Party[i]);}}if(ccount!=0)printf("\n");return 0;
}