poj6029

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
题意：

有n个点，有下面两种操作
从当前点向前面所有点连一条边
从当前点不向任何点连边
问构成的图是不是一个二分图。

做这一题首先了解一下什么是二分图。
二分图：无向图G为二分图的充分必要条件是，G至少有两个顶点，且其所有回路的长度均为偶数
#include <bits/stdc++.h>

using namespace std;

const int N = 100000 + 10;

int a[N];

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

    return 0;
}