题目链接：https://vjudge.net/problem/POJ-2236

题目大意：有一个计算机网络的所有线路都坏了，网络中有n台计算机，现在你可以做两种操作，修理（O）和检测两台计算机是否连通（S），只有修理好的计算机才能连通。连通有个规则，两台计算机的距离不能超过给定的最大距离D（一开始会给你n台计算机的坐标）。检测的时候输出两台计算机是否能连通。

只要每次修好一台计算机的同时把周围距离小于等于d的计算机加入集合就行

#include<set>#include<map>#include<stack>#include<queue>#include<cmath>#include<cstdio>#include<cctype>#include<string>#include<vector>#include<climits>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define endl ‘\n‘#define max(a, b) (a > b ? a : b)#define min(a, b) (a < b ? a : b)#define zero(a) memset(a, 0, sizeof(a))#define INF(a) fill(a, a+maxn, INF);#define IOS ios::sync_with_stdio(false)#define _test printf("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n")using namespace std;typedef long long ll;typedef pair<int, int> P;typedef pair<double, int> P2;const double pi = acos(-1.0);const double eps = 1e-7;const ll MOD = 1000000007LL;const int INF = 0x3f3f3f3f;const int _NAN = -0x3f3f3f3f;const double EULC = 0.5772156649015328;const int NIL = -1;template<typename T> void read(T &x){ x = 0;char ch = getchar();ll f = 1; while(!isdigit(ch)){if(ch == ‘-‘)f*=-1;ch=getchar();} while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;}const int maxn = 1e3+10;int p[maxn], vis[maxn];P cp[maxn];int __find(int root) { int son = root, temp; while(root != p[root]) root = p[root]; while(son != root) { temp = p[son]; p[son] = root; son = temp; } return root;}void merge(int son1, int son2) { p[__find(son1)] = p[__find(son2)];}int main(void) { IOS; int n, d; while(cin >> n >> d) { zero(vis); for (int i = 0; i<n; ++i) p[i] = i; for (int i = 0; i<n; ++i) cin >> cp[i].first >> cp[i].second; string s; while(cin >> s) { if (s[0] == ‘O‘) { int pos; cin >> pos; vis[pos-1] = true; for (int i = 0; i<n; ++i) if (i != pos-1 && vis[i] && (cp[i].first-cp[pos-1].first)*(cp[i].first-cp[pos-1].first) + (cp[i].second-cp[pos-1].second)*(cp[i].second-cp[pos-1].second) <= d*d) merge(i, pos-1); } else if (s[0] == ‘S‘) { int a, b; cin >> a >> b; if (__find(a-1) == __find(b-1)) printf("SUCCESS\n"); else printf("FAIL\n"); } } } return 0;}