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.
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.
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
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[];
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
count=;
for(i=; i<n; i++)
{
scanf("%d",&a[i]);
}
if(n%==)
{
printf("No\n");///奇数不可能配对
}
else
{
for(i=; i<n; i++)
{
if(a[i]==)
{
if(count==)
{
count=;
}
else
{
count--;
}
}
else
{
count++;
}
}
if(count==)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
}
return ;
}
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=;
for(i=; i<n; i++)
{
scanf("%d",&a);
if(a==||count==)
{
count++;
}
else
{
count--;
}
}
if(count==)
{
printf("Yes\n");
}
else
{
printf("No\n");
} }
return ;
}
思路本质上是一样的。。。。。。