HDU 6029（思维）

传送门

题面：

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1220    Accepted Submission(s): 553

Problem Description

Little Q loves playing with different kinds of graphs very much. One day he thought about an interesting category of graphs called ``Cool Graph'', which are generated in the following way:
Let the set of vertices be {1, 2, 3, ..., n}. You have to consider every vertice from left to right (i.e. from vertice 2 to n). At vertice i, you must make one of the following two decisions:
(1) Add edges between this vertex and all the previous vertices (i.e. from vertex 1 to i−1).
(2) Not add any edge between this vertex and any of the previous vertices.
In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices. A perfect matching is a matching that each vertice is covered by an edge in the set.
Now Little Q is interested in checking whether a ''Cool Graph'' has perfect matching. Please write a program to help him.

Input

The first line of the input contains an integer T(1≤T≤50), denoting the number of test cases.
In each test case, there is an integer n(2≤n≤100000) in the first line, denoting the number of vertices of the graph.
The following line contains n−1 integers a2,a3,...,an(1≤ai≤2), denoting the decision on each vertice.

Output

For each test case, output a string in the first line. If the graph has perfect matching, output ''Yes'', otherwise output ''No''.

Sample Input

3 2 1 2 2 4 1 1 2

Sample Output

Yes No No

Source

2017中国大学生程序设计竞赛 - 女生专场

题目描述：一种所有结点都有边与之相连的匹配叫做完美匹配。现在你有N个结点，对于n-1个结点，你有两种操作：

(1):将这个结点与之前的所有结点都连一条边

(2):不进行操作；

问你组成的这张图有没有可能是完美匹配。

题面分析：这是一道很有意思的思维题。首先要明确的一点是题目只要我们求是否可能是完美匹配，而不是让我们判断是否一定是完美匹配。对于每个结点，当我们要进行第一种操作时，即意味着我们这个结点可以与前面的任意一个结点进行匹配；而当我们进行第二种操作的时候，意味着这个结点是完全孤立的。因为每次操作1时，当前节点的匹配是任意的，考虑到这点，我们可以考虑使用队列去做，即当进行操作1的时候，队列深度减1，当操作为2时，将队列深度加1，最后判断队列是否非空即可。

而又考虑到这题中的结点对结果没有影响，故直接开一个数进行加一减一的模拟即可。

#include <bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){int n,num;cin>>n;if(n%2==1){for(int i=1;i<n;i++){cin>>num;}puts("No");continue;}int que=1;for(int i=1;i<n;i++){cin>>num;if((num==1&&que==0)||num==2){que++;}else que--;}if(que==0){puts("Yes");}else puts("No");}return 0;
}