Couples 夫妻配对

Description

N couples are standing in a circle, numbered consecutively clockwise from 1 to 2N. Husband and wife do not always stand together. We remove the couples who stand together until the circle is empty or we can't remove a couple any more.

Can we remove all the couples out of the circle?

Input

There may be several test cases in the input file. In each case, the first line is an integer N(1 <= N <= 100000)----the number of couples. In the following N lines, each line contains two integers ---- the numbers of each couple.
N = 0 indicates the end of the input.

Output

Output "Yes" if we can remove all the couples out of the circle. Otherwise, output "No".

Sample Input

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41 42 35 67 821 32 40

Sample Output

YesNo

Problem Source: ZSUACM Team Member

我的解法：

#include <stack>
#include <iostream>
#include <vector>

using namespace std;

int main()
{

int n;         //共有几对夫妻
while(cin>>n){   //读取每组测试数据中夫妻的对数
if(n==0) return 0;  //读到0测试数据结束
int a,b;           //接受数据
vector<int> cou;    //接受夫妻数据向量，其中向量下标为夫妻的序号减一
stack<int> cirl;   //用栈来验证是否都符合要求
vector<int>::iterator it;
for(int z=0;z<n*2;z++) //初始化向量
{
cou.push_back(z);
}

for(int x=0;x<n;x++)  //读取每对夫妻
{
cin>>a>>b;        
a=a-1;
b=b-1;
cou[a]=x;       //每对夫妻在向量中的数值相同
cou[b]=x;
}

for(it=cou.begin();it!=cou.end();it++)
{
if(!cirl.empty())
{
if(cirl.top()==(*it))  //数值相同，为夫妻，栈顶数据移除
{
cirl.pop();       
}
else cirl.push(*it);   //如果不相同，则入栈，查找下一个
}
else cirl.push(*it); 
}
if(!cirl.empty())           //栈为空则符合题目要求
cout<<"No"<<endl;
else cout<<"Yes"<<endl;
}
return 0;
}