HDU6029(思维题）

Graph Theory

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 257 Accepted Submission(s): 130

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

解题思路：从最后一个数开始，如果值为2，则ans++（可以与前面匹配的数量），如果为1，则需要与后面的匹配，则只需要ans–，（用掉一个匹配），如果ans < 0，则no，

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int n;
int a[maxn];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        a[1] = 2;
        for(int i = 2; i <= n; i++)
        {
            scanf("%d",&a[i]);
        }
        if(n&1) printf("No\n");
        else
        {
            int ans = 0;
            bool flag = true;
            for(int i = n; i >= 1; i--)
            {
                if(a[i] == 1) ans++;
                else
                {
                    ans--;
                    if(ans < 0)
                    {
                        flag = false;
                        break;
                    }
                }
            }
            if(flag) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}