HDU6029 Graph Theory

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 477    Accepted Submission(s): 229

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

3212241 1 2

 

Sample Output

YesNoNo

 

题意：给n个顶点，从第一个点开始操作，每个点有两种操作：1、将当前结点和之前的所有结点都加一条边 2、当前结点与之前的所有结点都不加边。问是否能够完美匹配？完美匹配是指所有的结点都有边连接，并且这些边中没有公共的顶点。

解析：用一个cnt来记录到当前结点时，之前的结点还未匹配的个数。如果当前结点操作是加边并且cnt>0，那么当前结点就和之前的一个结点匹配，使cnt--；否则cnt++。直到最后判断cnt是否为0，为0说明所有点都匹配了。

代码：

#include <stdio.h>int main(){    int t, n, cnt, x;    scanf("%d", &t);    while(t--){        scanf("%d", &n);        cnt = 1;        for(int i = 1; i < n; i++){            scanf("%d", &x);            if(cnt == 0)                cnt++;            else{                if(x == 1)                    cnt--;                else                    cnt++;            }        }        if(cnt != 0)            puts("No");        else                puts("Yes");    }    return 0;}