hdu6029 Graph Theory (2017女生赛)

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 167    Accepted Submission(s): 84

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

 

再说题意之前我要先学个单词：

所以说 a set of 的意思是 一组...

只需找到的是一组边使得满足perfect matching！

题意：有n个点让你进行操作，从第二点开始，你可以开始操作，并且只能操作其中的任意一个，输入给出了点数和各点的操作，现在要求的就是是否有一组边满足perfect matching；perfect matching 的意思就是任何一个点当且仅当被一条边所覆盖；

两种操作分别是，1：当前点和所有前面的点都要连；2：当前点与前面任意点都不去相连！

思路：完美匹配就是： 。——。 。——。...... 这种方式；所以点数必须为偶数个。第一个点没有操作，可以假设它的操作为2，然后后面遇到一个1就和前面的2匹配一下，作何如果都匹配成功，就ok啦！

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100005;
typedef long long ll;
const ll tomod=1e9+7;

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,x;
        int flag2=1;
        scanf("%d",&n);

        for(int i=2;i<=n;i++)
        {
            scanf("%d",&x);
            if(x==1)
            {
                if(flag2>0)
                    flag2--;
                else
                    flag2++;
            }
            else if(x==2)
                flag2++;
        }

        if(flag2!=0||n%2!=0)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}