Graph Theory

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 $T(1\leq T\leq50)$, denoting the number of test cases.
In each test case, there is an integer $n(2\leq n\leq 100000)$ in the first line, denoting the number of vertices of the graph.
The following line contains $n-1$ integers $a_2,a_3,...,a_n(1\leq a_i\leq 2)$, 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

4

1 1 2

Sample Output

Yes

No

题目意思：有n个点，这里给出了n-1个数（第0个点没有操作，所以不用）表示每个点的操作状态，操作1表示当前点与之前出现的所有的点连成一条边，操作2代表什么也不做，问最后是否每一个点都有一个点与其配对（两两配对）。

解题思路：英语水平确实太差了，上来看到graph，以为是图论，因为图论的内容没有学习，很打怵，不过榜单上和我水平差不多的队友有做出来的，就明白这不是一道难题，其实这应该算是一道找规律的题，我们很容易知道当n为奇数的时候是不可能出现两两匹配的。当n为偶数时，用count表示前面有多少个未配对的点，如果前面有未配对的点则，若操作为1,，则count--，若操作为2则count++。如果前面所有的点都匹对成功则，若操作为1,，则count=1（因为前面没有点与其配对），若操作为2则count++，最后如果count=0，则说明完美匹配perfect matching。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int t,n,i,count;
 8     int a[100010];
 9     scanf("%d",&t);
10     while(t--)
11     {
12         scanf("%d",&n);
13         count=1;
14         for(i=1; i<n; i++)
15         {
16             scanf("%d",&a[i]);
17         }
18         if(n%2==1)
19         {
20             printf("No\n");///奇数不可能配对
21         }
22         else
23         {
24             for(i=1; i<n; i++)
25             {
26                 if(a[i]==1)
27                 {
28                     if(count==0)
29                     {
30                         count=1;
31                     }
32                     else
33                     {
34                         count--;
35                     }
36                 }
37                 else
38                 {
39                     count++;
40                 }
41             }
42             if(count==0)
43             {
44                 printf("Yes\n");
45             }
46             else
47             {
48                 printf("No\n");
49             }
50         }
51     }
52     return 0;
53 }

看见有大佬写出了这样很简单的代码，我也学习一下：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 using namespace std;
 5 int main()
 6 {
 7     int t,n,i,j,a,count;
 8     scanf("%d",&t);
 9     while(t--)
10     {
11         scanf("%d",&n);
12         count=1;
13         for(i=1; i<n; i++)
14         {
15             scanf("%d",&a);
16             if(a==2||count==0)
17             {
18                 count++;
19             }
20             else
21             {
22                 count--;
23             }
24         }
25         if(count==0)
26         {
27             printf("Yes\n");
28         }
29         else
30         {
31             printf("No\n");
32         }
33 
34     }
35     return 0;
36 }

思路本质上是一样的。。。。。。

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