HDU6029-Graph Theory

Graph Theory

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
                                                                                                             Total Submission(s): 0    Accepted Submission(s): 0

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

 

题意：有n个点，每个点有两种连边方式，1表示和前面所有的点连边，2表示和前面所有的点不连边，问能不能构成完美匹配

解题思路：若n为奇数，那么必然不能构成完美匹配，n为偶数时，从后向前判，看方式1的个数是否一直大于方式2

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100005];

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