(PAT 1121) Damn Single (哈希)

灰太狼 2022-03-20 11:28 173阅读 0赞

"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 (≤ 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.

### 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

解题思路:

利用哈希思想解决,创建三个哈希表,一个用来保存此人得配偶,一个用来保存此人是否已婚,最后一个用来保存此人是否在派对上

判断是否单身时,先判断此人是否已婚,如果已婚,判断此人得配偶是否在派对上,查表即可

#include <iostream>
    #include <algorithm>
    #include <vector>
    using namespace std;
    const int MAXN = 100010;
    int N, M;
    int marrigeHash[MAXN];
    bool isMArrage[MAXN] = { false };
    bool atParty[MAXN] = { false };
    int guestesOnline[MAXN];
    vector<int> singleList;
    int main() {
    	//初始化
    	fill(marrigeHash, marrigeHash + MAXN, -1);
    	cin >> N;
    	int couple1, couple2;
    	for (int i = 0; i < N; ++i){
    		cin >> couple1 >> couple2;
    		marrigeHash[couple1] = couple2;
    		marrigeHash[couple2] = couple1;
    		isMArrage[couple1] = true;
    		isMArrage[couple2] = true;
    	}
    	cin >> M;
    	for (int i = 0; i < M; ++i) {
    		cin >> guestesOnline[i];
    		atParty[guestesOnline[i]] = true;
    	}
    	for (int i = 0; i < M; ++i) {
    		if (!isMArrage[guestesOnline[i]]) {
    			singleList.push_back(guestesOnline[i]);
    		}
    		else {
    			if (!atParty[marrigeHash[guestesOnline[i]]]) {
    				singleList.push_back(guestesOnline[i]);
    			}
    		}
    	}
    	sort(singleList.begin(), singleList.end());
    	cout << singleList.size() << endl;
    	for (int i = 0; i < singleList.size(); ++i) {
    		printf("%05d", singleList[i]);
    		if (i < singleList.size() - 1) cout << " ";
    	}
    	system("PAUSE");
    	return 0;
    }