Wireless Network POJ-2236

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

题目大意

地震了有n个电脑要修，两个电脑要连接距离不能超过d。如果AB要连接有两种情况，A直接连接B， AB不能直接连接但是A连接C，B连接C。给你两种操作，O意味着修复S意味着测试能否连接

分析

很明显是一个并查集题，模版加上一些限制就能过。

代码

#include <cstdio>#include <iostream>#include <string.h>#include <cmath>#include <cstdlib>using namespace std;#define NUM 1001int pre[NUM];int pos[NUM][NUM];bool conn[NUM];float dis[NUM];int n, d;int find_(int x){    int t = x;    while(t != pre[t]){        t = pre[t];    }    int k = x;    while(k != t){        int s = pre[k];        pre[k] = t;        k = s;    }    return t;}bool join_(int x, int y){    int xx = find_(x);    int yy = find_(y);    if(xx != yy){        pre[yy] = xx;        return false;    }    return true;}void op_0(int p){    conn[p] = true;    for(int i = 1; i <= n; i++){        if(conn[i] && conn[p]){                int &x1 = pos[i][0];                int &y1 = pos[i][1];                int &x2 = pos[p][0];                int &y2 = pos[p][1];                int dx = abs(x1 - x2);                int dy = abs(y1 - y2);                float dd;                dd = sqrt(dx * dx + dy * dy);                if(dd <= d){                    join_(i, p);                }            }        }}bool op_S(int p, int q){    int pp = find_(p);    int qq = find_(q);    if(pp == qq){        return true;    }    else{        return false;    }}int main(){//  freopen("B.txt", "r", stdin);    scanf("%d%d", &n, &d);    for(int i = 0; i <= n; i++){        pre[i] = i;    }    memset(conn, false, sizeof(conn));    for(int i = 1; i <= n; i++){        int a, b;        scanf("%d%d", &a, &b);        pos[i][0] = a;        pos[i][1] = b;    }    char s;    while(cin >> s){        if(s == 'O'){                int p;                scanf("%d", &p);                op_0(p);        }        else if(s == 'S'){                int pp, q;                scanf("%d%d", &pp, &q);                if(op_S(pp, q)){                    printf("SUCCESS\n");                }                else{                    printf("FAIL\n");                }               }           }    return 0;}