hdu 6029 Graph Theory 【直接连线】

hdu 6029 Graph Theory 【直接连线】

对于输入的a等于1  如果前面有空闲的点 

直接匹配 否则自己变成空闲的点

a=2 的时候  空闲的点加一 

直到最后 如果没有空余的点 输出yes

Graph Theory


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 145    Accepted Submission(s): 76




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

#include <stdio.h>#include <bits/stdc++.h>#define mod 1000000007#define read(); freopen("input.txt","r",stdin);typedef long long ll;using namespace std;int t,n;int main(){cin>>t;while(t--){cin>>n;int a;int sum=1;for(int i=1;i<n;i++){scanf("%d",&a);if(a==1){if(sum>=1)sum--;else sum++;}else{sum++;}}if(sum==0)printf("Yes\n");else printf("No\n");}return 0;}