hdu6029 Graph Theory (2017女生赛)

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 167    Accepted Submission(s): 84

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 toi−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 edgein 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

3212241 1 2

 

Sample Output

YesNoNo

 

Source

2017中国大学生程序设计竞赛 - 女生专场

 

再说题意之前我要先学个单词：

所以说a set of 的意思是 一组...

只需找到的是一组边使得满足perfect matching！

题意：有n个点让你进行操作，从第二点开始，你可以开始操作，并且只能操作其中的任意一个，输入给出了点数和各点的操作，现在要求的就是是否有一组边满足perfect matching；perfect matching 的意思就是任何一个点当且仅当被一条边所覆盖；

两种操作分别是，1：当前点和所有前面的点都要连；2：当前点与前面任意点都不去相连！

思路：完美匹配就是： 。——。 。——。...... 这种方式；所以点数必须为偶数个。第一个点没有操作，可以假设它的操作为2，然后后面遇到一个1就和前面的2匹配一下，作何如果都匹配成功，就ok啦！

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100005;typedef long long ll;const ll tomod=1e9+7;int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,x;        int flag2=1;        scanf("%d",&n);        for(int i=2;i<=n;i++)        {            scanf("%d",&x);            if(x==1)            {                if(flag2>0)                    flag2--;                else                    flag2++;            }            else if(x==2)                flag2++;        }        if(flag2!=0||n%2!=0)            printf("No\n");        else            printf("Yes\n");    }    return 0;}