Graph Theory

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 872    Accepted Submission(s): 415


Problem DescriptionLittle 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≤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.
OutputFor each test case, output a string in the first line. If the graph has perfect matching, output ''Yes'', otherwise output ''No''.

Sample Input3212241 1 2Sample OutputYesNoNo

題意：第一行輸入一個整數T

表示測試資料，後面輸入T組資料，一行輸入n表示有n個點。然後進行n-1個操作，1表示能和之前的所有點相連，2就是無操作，問是否能有Perfect match。完美匹配就是線與線無公共點，也就是兩兩相連，無第三條線與其相連。

思路：看懂題目是關鍵，看懂了之後，這道題就很簡單，但是題目極其的難懂，我是一臉懵逼的看完所有描述，我思前想後，比如一個樣例11 2 1 2 1，或許有人會和我之前一樣理解以為最後一個1必須和之前出現的所有數都連一遍，那第一個數就會和後面所有的1都連一遍，那怎麼可能會出現完美匹配。思前想後好久，明白了原來這個1是可以只連一次的，而此時就可能出現完美匹配。說穿了本題就是找兩兩相連，看能不能連完。如果後面出現了一個2，那麼2的後面必須出現一個1與其相連，否則就會有一個沒有相連，或者有公共點。這樣就是2必須對應1，而1可以對應1和2.不得不說題目模糊的一匹，難以理解，真叫人頭禿。

AC程式碼：

#include<bits/stdc++.h>
using namespace std;

#define maxn 100005
int main()
{
    int a[maxn];
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
         int sum=1;
        for(int i=1;i<n;i++)
        {
            if(sum==0) sum++;
            else
            {
                if(a[i]==1) sum--;
                else sum++;
            }
        }
        if(sum!=0) printf("No\n");
        else printf("Yes\n");
    }
}

新手玩家，有心臟類疾病，不喜勿噴。