hdu-6029 Graph Theory

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.

InputThe first line of the input contains an integer T(1≤T≤5, 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.OutputFor 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

OJ-ID：
hdu-6029
author：
Caution_X

date of submission：
20191017

tags：
思维

description modelling：
给定n个点，除第一个点外每一个点有两种操作：
操作1：选择一个比该点小的点匹配
操作2：该点之前的点不得与该点匹配
问：最终能否达到完美匹配

major steps to solve it：
(1).显然n为奇数不可能，当n为偶数时进入(2)
(2).从后往前遍历，保证操作1的个数始终大于操作2的个数即可

AC code:

#include
using namespace std;
int a[100005];
int main()
{
    //freopen("input.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--) {
        int n;
        scanf("%d",&n);
        for(int i=0;i1;i++) {
            scanf("%d",&a[i]);
        }
        if(n%2!=0) {
            printf("No\n");
            continue;
        }
        int p=0;
        bool f=false;
        for(int i=n-2;i>=0;i--) {
            if(a[i]==1)    p++;
            else p--;
            if(p<0) {
                f=true;
                break;
            }
        }
        if(!f)    printf("Yes\n");
        else printf("No\n");
    }
    return 0; 
}