洛谷 P2978 [USACO10JAN]下午茶时间Tea Time

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

输入样例#1： 复制

5 3 3
2 5
2 3
4 5
2 3
3 5
1 5

输出样例#1： 复制

Y
Y
N

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[];
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=;i<=n;i++)    fa[i]=i;
    for(int i=;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=;i<=k;i++){
        int x,y;
        scanf("%d%d",&x,&y);
        if(find(x)==find(y))    cout<<"Y"<<endl;
        else cout<<"N"<<endl;
    }
}