傳送門題面：

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 Input3 2 1 2 2 4 1 1 2
Sample OutputYes No No
Source 2017中國大學生程序設計競賽 - 女生專場

題目描述：一種所有結點都有邊與之相連的匹配叫做完美匹配?，F在你有N個結點，對于n-1個結點，你有兩種操作：

(1):將這個結點與之前的所有結點都連一條邊(2):不進行操作；

問你組成的這張圖有沒有可能是完美匹配。題面分析：這是一道很有意思的思維題。首先要明確的一點是題目只要我們求是否可能是完美匹配，而不是讓我們判斷是否一定是完美匹配。對于每個結點，當我們要進行第一種操作時，即意味著我們這個結點可以與前面的任意一個結點進行匹配；而當我們進行第二種操作的時候，意味著這個結點是完全孤立的。因為每次操作1時，當前節點的匹配是任意的，考慮到這點，我們可以考慮使用隊列去做，即當進行操作1的時候，隊列深度減1，當操作為2時，將隊列深度加1，最后判斷隊列是否非空即可。

而又考慮到這題中的結點對結果沒有影響，故直接開一個數進行加一減一的模擬即可。#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; }

轉載于:https://www.cnblogs.com/Chen-Jr/p/11007311.html

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