HDU 6029 Graph Theory

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1520    Accepted Submission(s): 642

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：输入n，表示有n个点（编号从1~n），对第2~n每个点都做且仅做一种处理（处理1：把这个点同这个点之前的每个点之间都加一条线；处理2：这个点不予任何点相连）。等处理完所有点之后问有没有符合完美图的。完美图的定义是存在子图，组图中的点两两配对，且没有单独存在的点。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,order;
        cin>>n;
        int flag=-1,sum=0;
        for(int i=2;i<=n;i++)
        {
            cin>>order;
            if(order==1)
            {
                if(sum<i-1)
                sum+=2;
            }///如果当前第i个点的操作是1，且当前加入集合sum的点的个数小于i-1，则加入第i-1个点和第i个点（i-1与i配对入集合sum）
        }
        if(n%2==1)printf("No\n");///奇数点一定存在被孤立的点，一定不符合
        else
        {
            if(sum==n) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}

方法二：记录其中孤立的点，当n为偶数且孤立的点个数为0时输出yes。（此法总体耗时较少）

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,order;
        scanf("%d",&n);
        int sum=1;///表示单独存在的点的个数
        for(int i=2;i<=n;i++)
        {
            scanf("%d",&order);
            if(order==1)///当对当前点i的操作为1，则会减去一个当前孤立的点，一同这个点i入集合
                sum--;
            else sum++;///当对i的操作为2时，记为孤立的点，孤立的点个数加一
            if(sum<0)///只有当前i点操作为1，但有没有孤立的点一直配对，才会出现sum<0的情况
                sum=1;///这时候把这个操作为1的点暂且记作孤立的点，等待下个操作为1的点与之配对。
           /// cout<<"sum="<<sum<<endl;
        }
        if(n%2==1)printf("No\n");
        else
        {///cout<<"*sum="<<sum<<endl;
            if(sum==0)printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}