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;
}