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  • HDU 6029 Graph Theory【水题】

    Graph Theory

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 448    Accepted Submission(s): 216


    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
     

    Source

    问题

    有n个点,1表示当前点与之前所有点连一条边,2表示不动,问通过选出其中的某些边,能否使所有的点都能够有一个点与其配对。

    思路:
    用cnt表示前面有多少个未配对的点,
    如果前面有未配对的点则,若操作为1,,则cnt--,若操作为2则cnt++
    如果前面所有的点都匹对成功则,若操作为1,,则cnt=1(因为前面没有点与其配对),若操作为2则cnt++

    #include<iostream>
    #include<algorithm>
    #include<cmath>
    #include<cstdio>
    #include<cstdlib>
    #include<queue>
    #include<map>
    #include<set>
    #include<stack>
    #include<bitset>
    #include<numeric>
    #include<vector>
    #include<string>
    #include<iterator>
    #include<cstring>
    #include<ctime>
    #include<functional>
    #define INF 0x3f3f3f3f
    #define ms(a,b) memset(a,b,sizeof(a))
    #define pi 3.14159265358979
    #define mod 1000000007
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    using namespace std;
    
    typedef pair<int, int> P;
    typedef long long ll;
    typedef unsigned long long ull;
    const int maxn = 100010;
    
    int n, a[maxn];
    
    int main()
    {
    	int t;
    	scanf("%d", &t);
    	while (t--)
    	{
    		int cnt = 1;
    		scanf("%d", &n);
    		for (int i = 2; i <= n; i++)
    		{
    			scanf("%d", a + i);
    		}
    		if (n % 2 != 0) puts("No");
    		else
    		{
    			for (int i = 2; i <= n; i++)
    			{
    				if (a[i] == 1)
    				{
    					if (cnt == 0) cnt = 1;
    					else cnt--;
    				}
    				else cnt++;
    			}
    			if (cnt > 0) puts("No");
    			else puts("Yes");
    		}
    	}
    }



    Fighting~
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  • 原文地址:https://www.cnblogs.com/Archger/p/8451647.html
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