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 $T(1leq Tleq50)$, denoting the number of test cases. 
In each test case, there is an integer $n(2leq nleq 100000)$ in the first line, denoting the number of vertices of the graph. 
The following line contains $n-1$ integers $a_2,a_3,...,a_n(1leq a_ileq 2)$, 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

 

 

 

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。

 

 

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
    int t,n,i,count;
    int a[100010];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        count=1;
        for(i=1; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        if(n%2==1)
        {
            printf("No
");///奇数不可能配对
        }
        else
        {
            for(i=1; i<n; i++)
            {
                if(a[i]==1)
                {
                    if(count==0)
                    {
                        count=1;
                    }
                    else
                    {
                        count--;
                    }
                }
                else
                {
                    count++;
                }
            }
            if(count==0)
            {
                printf("Yes
");
            }
            else
            {
                printf("No
");
            }
        }
    }
    return 0;
}

 看见有大佬写出了这样很简单的代码，我也学习一下：

 

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
    int t,n,i,j,a,count;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        count=1;
        for(i=1; i<n; i++)
        {
            scanf("%d",&a);
            if(a==2||count==0)
            {
                count++;
            }
            else
            {
                count--;
            }
        }
        if(count==0)
        {
            printf("Yes
");
        }
        else
        {
            printf("No
");
        }

    }
    return 0;
}

 思路本质上是一样的。。。。。。