Graph Theory

标签：names   ==   continue   isp   ase   limit   操作   perfect   element   

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

Total Submission(s): 140    Accepted Submission(s): 74

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

Sample Output

Yes

No

No

 

// 题意很难懂啊。。。一个 1-n 长，为 n 的一排点，对于每一个点，从左到右，有两种操作中的一个 (1)对于之前的点，每个连条边 (2)什么都不做

问是否存在一个边的集合，所有点两个一组相连，没有孤立点。

题解:很奇怪的一个题，就是不太懂题意，懂了的话还是很简单的。

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 using namespace std;
 5 #define MX 100005
 6 
 7 int n;
 8 int data[MX];
 9 
10 int main()
11 {
12     int T;
13     cin>>T;
14     while (T--)
15     {
16         scanf("%d",&n);
17         for (int i=2;i<=n;i++)
18             scanf("%d",&data[i]);
19         if (n%2)
20         {
21             printf("No\n");
22             continue;
23         }
24         int ok = 1;
25         int cnt = 0;
26         for (int i=n;i>=2;i--)
27         {
28             if (data[i]==1) cnt++;
29             else cnt--;
30             if (cnt<0)
31             {
32                 ok=0;
33                 break;
34             }
35         }
36         if (ok)
37             printf("Yes\n");
38         else
39             printf("No\n");
40     }
41     return 0;
42 }

View Code

 

 

 

Graph Theory

标签：names   ==   continue   isp   ase   limit   操作   perfect   element   

原文地址：http://www.cnblogs.com/haoabcd2010/p/6827856.html