# PAT Advanced 1118 Birds in Forest

PAT Advanced 1118 Birds in Forest tests Union-Find algorithm, the maximum number of trees is the number of disjoint-set as two disjoint-set can possibly merged again.

## PAT Advenced 1118 Birds in Forest

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

## PAT Advenced 1118 Birds in Forest solution

Similar to PAT 1114 Family Property problem, we can use the Disjoit Set data structure to represent the trees, one tree corresponds to a set, merge two sets when we see any two birds are in the same picture, the finaly number of trees is not the exact number of the trees in the forest, but is the uppder bound, that is, the maximum number of trees, because given more pictures, existing sets could be merged again, and the number of trees could only be decreasing.

The following is the C++ source code that could be accepted by PAT OJ to pass this PAT Advenced 1118 Birds in Forest:

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#include <cstdio>
#include <set>
#include <algorithm>
using std::set;
using std::max_element;
int p[10005];
void init() {
for (int i = 0; i < 10005; ++i) {
p[i] = i;
}
}
int find(int x) {
return p[x] == x ? x : (p[x] = find(p[x]));
}
void setUnion(int i, int j) {
p[find(i)] = p[find(j)];
}
bool isUnion(int i, int j) {
return find(i) == find(j);
}
int main() {
int n, k, b, bird;
init();
set<int> idx;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", &k);
if (k > 0) {
scanf("%d", &bird);
idx.insert(bird);
}
for (int j = 1; j < k; ++j) {
scanf("%d", &b);
setUnion(bird, b);
idx.insert(b);
}
}
set<int> f;
for (const auto &e: indexes) {
f.insert(find(e));
}
printf("%d %d\n", f.size(), *max_element(idx.begin(), idx.end()));
int q, tt, kk;
scanf("%d", &q);
for (int i = 0; i < q; ++i) {
scanf("%d %d", &tt, &kk);
if (isUnion(tt, kk)) {
printf("Yes\n");
} else {
printf("No\n");
}
}
return 0;
}

Note that do not print out YES instead of Yes, NO instead of No.

## Summary

PAT Advanced 1118 Birds in Forest tests Union-Find algorithm, the maximum number of trees is the number of disjoint-set as two disjoint-set can possibly merged again.