A circle of friends is a network of friend relationships. If A is a friend of B, then B is considered a friend of A no matter B admits or not, and they are said to belong to the same circle. Here we assume that friendship is transitive, that is, if A is a friend of B, and B is a friend of C, then A is a friend of C and the three of them all belong to the same circle.
On the other hand, A is not so close to C as B is. We define the distance D(X, Y) between two friends X and Y as the minimum number of friends between them. For example, D(A, B) = 0, and D(C, A) = 1. The diameter of a friends circle is the maximum distance between any pair of friends in the circle.
Now given some people's relationships, you are supposed to find the number of friends circles and the circle with the largest diameter.
Input Specification:
Each input file contains one test case. For each case, the first line gives an integer N (2≤N≤1000), which is the total number of people involved, and hence they are numbered from 1 to N. Then N lines follow, each in the format:
k p1 ... pk
where k (0≤k<min(10,N)) is the number of friends and p1 to pk (if k>0) are the friends' indices. The i-th line corresponds to the i-th person. All the numbers in a line are separated by spaces. It is guaranteed that no one is given as a friend of oneself.
Output Specification:
For each case, print in a line the number of friends circles, and the largest diameter, separated by exactly one space.
Sample Input:
17
2 15 12
1 17
2 16 9
1 8
4 10 13 15 14
0
2 11 14
1 4
2 2 3
2 13 11
2 15 7
2 1 14
2 5 15
0
0
1 3
1 2
Sample Output:
4 3
題目大意：劃定朋友圈，只要A，B之間可能通過好友線聯繫起來，就是一個圈子的。現在要求有多少個朋友圈。另外A,B之間需要最短通過幾個好友才能聯繫上爲A,B之間的距離，現在還要求出這些人之間最大的距離。
解題思路：乍一看像是並查集的題目。但是對於距離的求解並查集就不起作用了，還得老老實實用BFS（理論上DFS也可以，不知道會不會超時，有興趣的可以嘗試一下）。本題思路可以參考甲級題目1021。
這裏思路是類似的，第一遍遍歷的時候存儲一下距離最大的那幾個點，再次進行遍歷的時候只要對這幾個點進行BFS，計算與他們相距最大的距離即可！
代碼：
#include<bits/stdc++.h>
using namespace std;
vector<set<int>> v;
set<int> lastlevel, anslevel;
bool visit[1005] = {0}, tempvisit[1005];
int n, k, temp, cnt = 0, ans = 0;
void BFS(int index){
queue<int> q;
vector<int> dis(n+1, 0);
q.push(index);
fill(tempvisit, tempvisit+1005, false);
visit[index] = true;
while(!q.empty()){
temp = q.front();
visit[temp] = true;
tempvisit[temp] = true;
q.pop();
if(dis[temp]-1 > ans){
lastlevel.clear();
lastlevel.insert(temp);
ans = dis[temp]-1;
}
else if(dis[temp] - 1 == ans)
lastlevel.insert(temp);
for(auto it = v[temp].begin(); it != v[temp].end(); it ++){
if(!tempvisit[*it]){
q.push(*it);
tempvisit[*it] = true;
dis[*it] = dis[temp]+1;
}
}
}
}
int main(){
scanf("%d", &n);
v.resize(n+1);
for(int i = 1; i <= n; ++ i){
scanf("%d", &k);
for(int j = 0; j < k; ++ j){
scanf("%d", &temp);
v[i].insert(temp);
v[temp].insert(i);
}
}
for(int i = 1; i <= n; ++ i){
if(visit[i])
continue;
++ cnt;
BFS(i);
}
anslevel = lastlevel;
fill(visit, visit+1005, false);
for(auto it = anslevel.begin(); it != anslevel.end(); ++ it)
BFS(*it);
printf("%d %d", cnt, ans);
}