P2978 [USACO10JAN]下午茶时间Tea Time
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
5 3 3 2 5 2 3 4 5 2 3 3 5 1 5
Y Y N
说明
感谢@蒟蒻orz神犇 提供翻译。
并查集模板题。不多加赘述，包括了路径压缩、寻找代表元素、集合合并，其他的操作都很好想。看懂这道题并查集就会了（会不会做题还请移步另一道题：小胖的奇偶）。
#include#include #include #include const int INF = 9999999;const int MAXN = 2000 + 10;const int MAXM = 1000 + 10;int m,n,q;int fa[MAXN];int find(int x){ return fa[x] == x? fa[x] : fa[x] = find(fa[x]);}int main(){ scanf("%d%d%d", &n, &m, &q); for(int i = 1;i <= n; i++) { fa[i] = i; } for(int i = 1;i <= m;i++) { int a,b; scanf("%d%d", &a, &b); fa[find(a)] = find(b); } for(int i = 1;i <= q; i++) { int a,b; scanf("%d%d", &a, &b); if(find(a) == find(b)) { printf("Y\n"); } else { printf("N\n"); } } return 0;}