题目描述
N (1 <= N <= 1000) cows, conveniently numbered 1..N all attend a tea time every day. M (1 <= M <= 2,000) unique pairs of those cows have already met before the first tea time. Pair i of these cows who have met is specified by two differing integers A_i and B_i (1 <= A_i <= N; 1 <= B_i <= N). The input never indicates that cows have met each other more than once.
At tea time, any cow i and cow j who have met a mutual friend cow k will meet sometime during that tea time and thus expand their circle of known cows.
Determine whether Q (1 <= Q <= 100) pairs of cows have met after tea times are held for long enough that no new cow meetings are occurring. Query j consists of a pair of different cows X_j and Y_j (1 <= X_j <= N; 1 <= Y_j <= N).
For example, suppose that out of cows 1 through 5, we know that 2 has met 5, 2 has met 3, and 4 has met 5; see (a) below.
2---3 2---3 2---3
| | | /|
1 --> 1 | | --> 1 | X |
| | |/ |
4---5 4---5 4---5
(a) (b) (c)In the first tea time, cow 2 meets cow 4, and cow 3 meets cow 5; see (b) above. In the second tea time, cow 3 meets cow 4; see (c) above.
N(1 <= N <= 1000)头奶牛，编号为1..N，在参加一个喝茶时间活动。在喝茶时间活动开始之前，已经有M(1 <= M <= 2,000)对奶牛彼此认识（是朋友）。第i对彼此认识的奶牛通过两个不相同的整数Ai和Bi给定(1<= Ai <= N; 1 <= Bi <= N)。输入数据保证一对奶牛不会出现多次。 在喝茶时间活动中，如果奶牛i和奶牛j有一个相同的朋友奶牛k，那么他们会在某次的喝茶活动中去认识对方（成为朋友），从而扩大他们的社交圈。 请判断，在喝茶活动举办很久以后（直到没有新的奶牛彼此认识），Q(1 <= Q <= 100)对奶牛是否已经彼此认识。询问j包含一对不同的奶牛编号Xj和Yj(1 <= Xj <= N; 1 <= Yj <= N)。 例如，假设共有1..5头奶牛，我们知道2号认识5号，2号认识3号，而且4号认识5号；如下图(a)。
2---3 2---3 2---3
| | | /|
1 --> 1 | | --> 1 | X |
| | |/ |
4---5 4---5 4---5
(a) (b) (c)在某次的喝茶活动中，2号认识4号，3号认识5号；如上图(b)所示。接下来的喝茶活动中，3号认识4号，如上图(c)所示。
输入输出格式
输入格式：
-
Line 1: Three space-separated integers: N, M, and Q
-
Lines 2..M+1: Line i+1 contains two space-separated integers: A_i and B_i
- Lines M+2..M+Q+1: Line j+M+1 contains query j as two space-separated integers: X_j and Y_j
行1：三个空格隔开的整数：N，M，和Q
行2..M+1：第i+1行包含两个空格隔开的整数Ai和Bi
行M+2..M+Q+1：第j+M+1行包含两个空格隔开的整数Xj和Yj，表示询问j
输出格式：
- Lines 1..Q: Line j should be 'Y' if the cows in the jth query have met and 'N' if they have not met.
行1..Q：如果第j个询问的两头奶牛认识， 第j行输出“Y”。如果不认识，第j行输出“N”
输入输出样例
说明
感谢@蒟蒻orz神犇 提供翻译。
思路：并查集
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int n,m,k; int fa[1010]; int find(int x){ if(fa[x]==x) return fa[x]; else return fa[x]=find(fa[x]); } int main(){ scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=n;i++) fa[i]=i; for(int i=1;i<=m;i++){ int x,y; scanf("%d%d",&x,&y); int dx=find(x); int dy=find(y); if(dx==dy) continue; else fa[dx]=dy; } for(int i=1;i<=k;i++){ int x,y; scanf("%d%d",&x,&y); if(find(x)==find(y)) cout<<"Y"<<endl; else cout<<"N"<<endl; } }