Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 194    Accepted Submission(s): 102

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：首先输入一个t，是样例数，然后输入一个n，表示有n个结点，下面有n-1个数，都是1和2，表示从第2个点开始到第n个点的选择，如果是1，表示这个点和前面所有的点中间建立一条边，2则是无操作，例如4  1 1 2，点2与点1相连，点3与点1点2相连，点4无操作。最后求是否构成完美匹配，完美匹配就是指每个点都配上对。
这里只需要考虑是否每个点都成对了，多余点边可以忽略不影响，每个点都可以与之前所有的点匹配，每次新判断一个点时都记录匹配数并获取之前匹配剩余数可与之匹配，这道题只需判断1和2，处理遍历点数与之前的匹配数关系
#include<stdio.h>#include<iostream>#include<string.h>using namespace std;int a[100005];int main(){    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        //a[1]=2;        for(int i=2;i<=n;i++)        {            cin>>a[i];        }        if(n%2)        {            printf("No\n");            continue;        }        int flag=0,sum=0;        for(int i=n;i>=2;i--)        {            if(a[i]==1)            {                sum++;            }            else            {                if(sum==0)                {                    flag=1;                    break;                }                 sum--;            }        }        if(flag==0)        {            printf("Yes\n");        }        else            printf("No\n");    }}