文章标题

浅析Pat 甲级1121. Damn Single（单身狗

题目：
“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

分析：
1.couple之间的对应关系，可以选择二维数组或者map。选择数组的时候，行表示couple的个数，列有两个，一个为丈夫，一个为妻子。选择map是必须注意：一对couple（A,B),A为key必须对应B，同时B也应该为key,对应A.
2.查询是否单身时，可以按照夫妻数目进行第一重循环，丈夫和妻子同时在待查询的数组中时，把相应的夫妻删除。
3.通过二维数组和map可以做出来，但是超时！！！。因此不能这样做，选择哈希！！

分析2
1.这道题无非就是妻子与丈夫对应，丈夫与妻子对应。那么选择哈希，妻子中存丈夫，丈夫中存妻子。查询的数组也用哈希。
2.具体：开两个数组，一个存couple，一个存待查找的。

代码

#include <iostream>
#include <string>
#include <stdlib.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
using namespace std;
int main(int argc, char** argv) {
    const int MAX = 100000;
    string couple[MAX]; 
    string peek[MAX];
    int n;
    cin>>n;
    for (int i = 0; i < MAX; i++) {//初始化 
            couple[i]=" ";
            peek[i] =" ";
        }
    for (int i = 0; i < n; i++) {
            string s; cin>>s;
            string ss; cin>>ss;
            couple[atoi(s.c_str())] = ss;//对应关系存储 
            couple[atoi(ss.c_str())] = s;
        }
        int m;
        cin>>m;
        for (int i = 0; i < m; i++) {
            string s;
            cin>>s;
            peek[atoi(s.c_str())] = s;
        }
        int sum = 0;
        for (int i = 0; i < MAX; i++) {
            if(!(peek[i]==" ")){
                string s = couple[atoi(peek[i].c_str())];//根据couple查到另一方 
                if(!(s==" ")){
                    if(!(peek[atoi(s.c_str())]==" ")){//查找数组中有 
                        peek[i]=" ";//查到置为空 
                        peek[atoi(s.c_str())]=" ";//查到置为空
                        sum+=2;
                    }
                }
            }
        }
        cout<<m-sum<<endl;
        string st="";
        for (int i = 0; i < MAX; i++) {//输出 
            if(!(peek[i]==" ")){
                st.append(peek[i]+" ");
            }
        }
        cout<<st.substr(0,st.length()-1);
    return 0;
}