Graph Theory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1796 Accepted Submission(s):
750
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.
, 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
,a
3
,...,a
n
(1≤a
i
≤2)
, denoting the decision on each vertice.
If the graph has perfect matching, output ''Yes'', otherwise output
''No''.
2
1
2
2
4
1 1 2
No
No
several similar problems for you: 6286 6285 6284 6283 6282
#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#include<vector>
#define ll long long
#define inf 0x3f3f3f3f
#define mod 1000000007;
using namespace std;
int a[];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
int s1=;
int s2=;
int x;
bool f=;
for(int i=;i<=n;i++)
{//2只能靠后面的1把它连上边 scanf("%d",&x);
a[i]=x;
}
for(int i=n;i>=;i--)//倒着遍历，从后往前看的话，1的数量一定要比2多
{
if(a[i]==) s1++;
else s2++;
if(s2>s1)
{
f=;
break;
}
}
if(n%==||!f||x!=) printf("No\n");//奇数个点肯定不行
else printf("Yes\n");
}
return ;
}