hdu 6029 Graph Theory(贪心)

本文介绍了一种名为CoolGraph的特殊图结构,并提供了一个算法来判断这种图是否包含完美匹配。完美匹配是指图中每顶点恰好被一条边覆盖的情况。通过输入一系列决策,算法能够确定CoolGraph是否存在这样的匹配。

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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 i1).
(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(1T50), denoting the number of test cases.
In each test case, there is an integer n(2n100000) in the first line, denoting the number of vertices of the graph.
The following line contains n1 integers a2,a3,...,an(1ai2), 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;
}
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