1021. Couples 栈和数组的实现

1021. Couples

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

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 integerN(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

41 42 35 67 821 32 40

Sample Output

YesNo

Problem Source

ZSUACM Team Member

很简单的题啊啊啊，和符号匹配很像。

上代码

#include <iostream>#include <stack>using namespace std;int main() {int N;while(cin >> N, N != 0) {int couple[200005] = {0};stack<int> pair;//对夫妻设置同样的号码1~N（夫妻代表的号码作为各自下标号）for (int i = 1; i <= N; i++) {int a, b;cin>>a>>b;couple[a] = couple[b]= i;}for (int i = 1; i <= 2*N; i++) {if (pair.empty() || couple[i] != pair.top()) //若栈为空或夫妻不匹配，则入栈pair.push(couple[i]);elsepair.pop();//否则（夫妻匹配）出栈}if (pair.size() > 0)cout<<"No"<<endl;elsecout<<"Yes"<<endl;}//system("pause");return 0;}