并查集入门笔记

1.并查集三个操作 |------初始化（par[i]保存根节点的键值）

|-------合并（集合）

|-------查询（是否在同一集合）

2.注意退化的情况，会是复杂度增高，用rak[]数组记录树的高度，每次合并两个树的时候都把深度小的数合并到深度大的树上

3.注意每次使用并查集都要给并查集par[]初始化（void init(int n)）;

函数：

//用角标代表元素，par代表根节点的元素值

int par[MAX_N], rak[MAX_N];

//初始化n个元素

void init(int i) {

for(int i = 0; i < n; i++) {

par[i] = i;

rak[i] = 0;

}

}

//搜索树的根

int Find(int x) {

if(par[x] == x) {

return x;

} else {

return par[x] = Find(par[x]);//递归找根

}

}

//合并连个点

void unite(int x, int y) {

x = Find(x);

y = Find(y);

if(x == y) {

return;

} else {

if(rak[x] < rak[y]) {

par[x] = y;

} else {

par[y] = x;

if(rak[x] == rak[y]) rak[x]++;

}

}

}

//判断x和y是否属于同一个集合

bool same(int x, int y) {

return Find(x) == Find(y);

}

例题：

Description

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The
computers are repaired one by one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded
as the intermediary of the communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.

In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.

Input

The first line contains two integers N and d (1 <= N <= 1001, 0 <= d <= 20000). Here N is the number of computers, which are numbered from 1 to N, and D is the maximum distance two computers can communicate directly. In the next
N lines, each contains two integers xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following
two formats:

1. "O p" (1 <= p <= N), which means repairing computer p.

2. "S p q" (1 <= p, q <= N), which means testing whether computer p and q can communicate.

The input will not exceed 300000 lines.

Output

For each Testing operation, print "SUCCESS" if the two computers can communicate, or "FAIL" if not.

Sample Input

4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4

Sample Output

FAIL
SUCCESS

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
typedef pair<int, int> P;
//#define LOCAL
#define INF 0x3f3f3f3f
#define MAX_N 1005

struct computer {
int number;//编号为角标
P coor;//坐标（x， y）
} comp[MAX_N], comp2[MAX_N];
int par[MAX_N], rak[MAX_N];

//返回两个点的距离
double Distance(int x, int y) {
return sqrt(((double)x * (double)x) + ((double)y * (double)y));
}

//并查集初始化
void init(int n) {
for(int i = 0; i < n; i++) {
par[i] = i;
rak[i] = 0;
}
}
//查根
int Find(int x) {
if(par[x] == x) return x;
else {
return par[x] = Find(par[x]);
}
}
//合并两个集合
void unite(int x, int y) {
x = Find(x);
y = Find(y);
if( x == y ) return;

if(rak[x] < rak[y]) {
par[x] = y;
} else {
par[y] = x;
if(rak[x] == rak[y]) rak[x]++;
}
}
//是否属于一个集合
bool same(int x, int y) {
return Find(x) == Find(y);
}

int main()
{
#ifdef LOCAL
freopen("b:\\data.in.txt", "r", stdin);
#endif
int N;
double d;
scanf("%d%lf", &N, &d);
init(N);//忘记了初始化，这bug也是不行不行滴
for(int i = 1; i <= N; i++) {
scanf("%d%d", &comp[i].coor.first ,&comp[i].coor.second);
comp[i].number = i;
}

char c;
int a, b;
int cnt = 0;
while(cin >> c) {
if(c == 'O') {
cin >> a;
comp2[cnt].coor = comp[a].coor;
//            cout << comp2[cnt].coor.first << " " << comp2[cnt].coor.second << endl;
comp2[cnt++].number = comp[a].number;

//取修好的电脑的数量
if(cnt > 1) {
for(int i = 0; i < cnt - 1; i++) {
//                   cout << Distance(abs(comp[a].coor.first - comp2[i].coor.first), abs(comp[a].coor.second - comp2[i].coor.second)) << endl;
if(Distance(abs(comp[a].coor.first - comp2[i].coor.first), abs(comp[a].coor.second - comp2[i].coor.second)) <= d) {
//修好的两个电脑的距离小于最远距离,合并两个点
//                        cout << a << " " << comp2[i].number << endl;
unite(a, comp2[i].number);//把修好的，并且可以相连的电脑放在一个集合
}
}
}
}
else
{
//输入为S的时候
cin >> a >> b;//a电脑和b电脑相连吗
if(same(a, b))
cout << "SUCCESS" << endl;
else
cout << "FAIL" << endl;
}
}
return 0;
}