2017中国大学生程序设计竞赛 - 女生专场（Graph Theory）

Posted 2020-11-07 蔡军帅

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1796    Accepted Submission(s): 750

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

 

 

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题意：给n个顶点，从第一个点开始操作，每个点有两种操作：1、将当前结点和之前的所有结点都加一条边 2、当前结点与之前的所有结点都不加边。问是否能够完美匹配？完美匹配是指所有的结点都有边连接，并且这些边中没有公共的顶点。

 

每一个2后面必须至少有一个1，那么倒着遍历

 

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
int a[100005];
int main()
{
   int T;
   scanf("%d",&T);
   while(T--)
   {
       int n;
       scanf("%d",&n);
       int s1=0;
       int s2=0;
       int x;
       bool f=1;
       for(int i=2;i<=n;i++)
       {//2只能靠后面的1把它连上边

           scanf("%d",&x);
           a[i]=x;
       }
       for(int i=n;i>=2;i--)//倒着遍历，从后往前看的话，1的数量一定要比2多
       {
           if(a[i]==1) s1++;
           else s2++;
           if(s2>s1)
           {
               f=0;
               break;
           }
       }
       if(n%2==1||!f||x!=1) printf("No\n");//奇数个点肯定不行
       else printf("Yes\n");
   }
   return 0;
}