POJ 2236 Wireless Network

本题的关键点在与，给定一个修复好的u，谁是和它连通的呢？  答案需要遍历全部的n，找出存在于并查集里的i，并且 距离（i, u） <=  d。注意这里的u可能本来就已经修好，所以只需要find（i） == find（u） 就可以满足条件.

/*********************************************** * Author: fisty * Created Time: 2015/2/26 15:48:34 * File Name   : A.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)#define MAX_N 1010int n, d;int vis[MAX_N];struct Point{    int x;    int y;}p[MAX_N];int dist(int i, int j){    Point a = p[i];    Point b = p[j];    return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);}int par[MAX_N];void init(){    for(int i = 1;i <= n; i++){        par[i] = i;    }}int find(int x){    if(x == par[x]) return x;    return par[x] = find(par[x]);}void unio(int x, int y){    x = find(x);    y = find(y);    par[x] = y;}int main() {    //freopen("in.cpp", "r", stdin);    //cin.tie(0);    //ios::sync_with_stdio(false);    cin >> n >> d;    Memset(vis, 0);    FOR(i, 1, n+1){        cin >> p[i].x >> p[i].y;    }    init();    char s[10];    while(scanf("%s", s) != EOF){        if(s[0] == 'O'){            int u;            cin >> u;            if(!vis[u]){                vis[u] = 1;                int a = find(u);                for(int i = 1;i <= n; i++){                    if(vis[i] && i != u && (find(i) == a || dist(i, u) <= d*d))                        unio(i, u);                }            }        }else{            int u, v;            cin >> u >> v;            if(vis[u] && vis[v]){                if(find(u) == find(v)){                    cout << "SUCCESS" << endl;                }else{                    cout << "FAIL" << endl;                }            }else                 cout << "FAIL" << endl;        }    }    return 0;}