HDU 6029 Graph Theory【水题】

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 448    Accepted Submission(s): 216

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

问题：

有n个点，1表示当前点与之前所有点连一条边，2表示不动，问通过选出其中的某些边，能否使所有的点都能够有一个点与其配对。

思路：

用cnt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cnt--，若操作为2则cnt++

如果前面所有的点都匹对成功则，若操作为1,，则cnt=1（因为前面没有点与其配对），若操作为2则cnt++

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<bitset>
#include<numeric>
#include<vector>
#include<string>
#include<iterator>
#include<cstring>
#include<ctime>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))
#define pi 3.14159265358979
#define mod 1000000007
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;

typedef pair<int, int> P;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100010;

int n, a[maxn];

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int cnt = 1;
		scanf("%d", &n);
		for (int i = 2; i <= n; i++)
		{
			scanf("%d", a + i);
		}
		if (n % 2 != 0) puts("No");
		else
		{
			for (int i = 2; i <= n; i++)
			{
				if (a[i] == 1)
				{
					if (cnt == 0) cnt = 1;
					else cnt--;
				}
				else cnt++;
			}
			if (cnt > 0) puts("No");
			else puts("Yes");
		}
	}
}