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
题意：有n个点，每个点有两种连边方式，1表示和前面所有的点连边，2表示和前面所有的点不连边，问能不能构成完美匹配
解题思路：若n为奇数，那么必然不能构成完美匹配，n为偶数时，从后向前判，看方式1的个数是否一直大于方式2
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
int a[100005];
int main()
{
int t,n;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i=1; i<n; i++)
scanf("%d",&a[i]);
if(n%2)
{
printf("No\n");
continue;
}
int cnt=0;
int flag=0;
for(int i=n-1; i>=1; i--)
{
if(a[i]==1) cnt++;
else
{
if(cnt==0)
{
flag=1;
break;
}
cnt--;
}
}
if(flag) printf("No\n");
else printf("Yes\n");
}
return 0;
}
