Graph Theory
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1796 Accepted Submission(s): 750
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.
Let the set of vertices be {1, 2, 3, ..., n
(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 a
2
,a
3
,...,a
n
(1≤a
i
≤2)
, denoting the decision on each vertice.
In each test case, there is an integer n(2≤n≤100000)
The following line contains n−1
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
Source
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题意:给n个顶点,从第一个点开始操作,每个点有两种操作:1、将当前结点和之前的所有结点都加一条边 2、当前结点与之前的所有结点都不加边。问是否能够完美匹配?完美匹配是指所有的结点都有边连接,并且这些边中没有公共的顶点。
每一个2后面必须至少有一个1,那么倒着遍历
#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[100005]; int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); int s1=0; int s2=0; int x; bool f=1; for(int i=2;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>=2;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]==1) s1++; else s2++; if(s2>s1) { f=0; break; } } if(n%2==1||!f||x!=1) printf("No\n");//奇数个点肯定不行 else printf("Yes\n"); } return 0; }