2017女生赛 1007 Graph Theory【】

Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

3212241 1 2

 

Sample Output

YesNoNo
过得自己都稀里糊涂的==
#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<queue>#include<stack>#include<vector>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))const int M=1e5+10;const int inf=0x3f3f3f3f;int i,j,k,n,m;int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        int ans=0;        int mark=1;        for(int i=2;i<=n;i++){                int aa;            scanf("%d",&aa);            if(aa==1&&mark){                ans+=2;                mark--;            }            else mark++;        }        if(ans==n)printf("Yes\n");        else printf("No\n");    }    return 0;}