hdu 6029 Graph Theory(贪心)

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 758    Accepted Submission(s): 363

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

这题主要要准确理解题意

A perfect matching is a matching that each vertice is covered by（ an） edge in the set. 这句话表示 每个点只连一条边

Now Little Q is interested in checking whether a ''Cool Graph''（ has） perfect matching.这句话表示 包含完美匹配就可以

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<map>
#include <bits/stdc++.h>
using namespace std;
const int N =1e5+10;
typedef  long long LL;
const LL mod = 1e9+7;
int dp1[N],dp2[N],a[N];

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int cnt=1;
        for(int i=2;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]==2) cnt++;
            else
            {
                if(cnt==0) cnt++;
                else cnt--;
            }
        }
        if(n&1||cnt) puts("No");
        else puts("Yes");
    }
    return 0;
}