1118 Birds in Forest（附测试点3分析）

题目：

Some scientists took pictures of thousands of birds in a forest. Assume that all the birds appear in the same picture belong to the same tree. You are supposed to help the scientists to count the maximum number of trees in the forest, and for any pair of birds, tell if they are on the same tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (≤104) which is the number of pictures. Then N lines follow, each describes a picture in the format:

K B1​ B2​ ... BK​

where K is the number of birds in this picture, and Bi​'s are the indices of birds. It is guaranteed that the birds in all the pictures are numbered continuously from 1 to some number that is no more than 104.

After the pictures there is a positive number Q (≤104) which is the number of queries. Then Q lines follow, each contains the indices of two birds.

Output Specification:

For each test case, first output in a line the maximum possible number of trees and the number of birds. Then for each query, print in a line Yes if the two birds belong to the same tree, or No if not.

Sample Input:

4
3 10 1 2
2 3 4
4 1 5 7 8
3 9 6 4
2
10 5
3 7

 

Sample Output:

2 10
Yes
No

 

题目大意：一幅画里面的鸟为同一棵树上的，问有多少棵树和多少只鸟，以及对于两只鸟判断是否在同一个树上

 

易错：测试点3超时可能是因为在进行集合合并时没有进行路径压缩，具体的路径压缩代码如下：

把当前查询结点的路径上的所有结点的父亲都指向根节点

int findFather(int x){
  int a = x;
  while(x != father[x]){
    x = father[x];
  } 
  while(a != father[a]){
    int z = a;
    a = father[a];
    father[z] = x;
  }
  return x;
}

 

代码：（满分）

#include<stdio.h>
#include<iostream>
using namespace std;
int father[10005], flag[10005]={0};
int findFather(int x){
  int a = x;
  while(x != father[x]){
    x = father[x];
  } 
  while(a != father[a]){
    int z = a;
    a = father[a];
    father[z] = x;
  }
  return x;
}
void Union(int x, int y){
    int faX = findFather(x);
    int faY = findFather(y);
    if(faX != faY){
        father[faY] = faX;
    }
}
int main(){
    int n, m, mx = -1;
    scanf("%d", &n);
    for(int i = 1; i <= 10000; i++){
        father[i] = i;
    }
    for(int i = 0; i < n; i++){
        int num, pre;
        scanf("%d", &num);
        for(int j = 0; j < num; j++){
            int index;
            scanf("%d", &index);
            if(index > mx) mx = index;
            if(j != 0) Union(pre, index);
            pre = index;
        }
    }
    int ans = 0;
    for(int i = 1; i <= mx; i++){
        if(father[i] == i){
            ans++;
        }
    }
    cout<<ans<<" "<<mx<<endl;
    scanf("%d", &m);
    for(int i = 0; i < m; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        int faX = findFather(x);
        int faY = findFather(y);
        if(faX != faY){
            printf("No\n");
        }else{
            printf("Yes\n");
        }
    }
}