Graph Theory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
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
过得自己都稀里糊涂的==
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b) memset(a,b,sizeof(a))
const int M=1e5+10;
const int inf=0x3f3f3f3f;
int i,j,k,n,m;
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
int ans=0;
int mark=1;
for(int i=2;i<=n;i++){
int aa;
scanf("%d",&aa);
if(aa==1&&mark){
ans+=2;
mark--;
}
else mark++;
}
if(ans==n)printf("Yes\n");
else printf("No\n");
}
return 0;
}
