HDU6029 Graph Theory 2017-05-07 19:04 40人阅读 评论(0) 收藏

Graph Theory

                                                                Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072
K (Java/Others)

                                                                                                             Total Submission(s): 17    Accepted Submission(s): 10

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

—————————————————————— ————————————

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，则不可能；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");
        }
        else
        {
            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;
}