poj2236Wireless Network之并查集解法

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Wireless Network
Time Limit: 10000MS Memory Limit: 65536K
Total Submissions: 29118 Accepted: 12049
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

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*分析:一看有几个关系组,还要求元素间关系的,果断并查集.
需要注意的是函数pow有三种重载形式：
long double pow(long double,int)
float pow(float,int)
double pow(double,int)
对于所给的参数a,b，编译器无法判断应该匹配哪个函数,因此需要强制转换一下,pow(double(a),b)*

/** Filename:    code.cpp* Created:     2017-07-25* Author:        wyl6 *[mail:[e-mail]]* Desciption:  Desciption*/#include <cstdio>#include <cstdlib>#include <iostream>#include <stack>#include <queue>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <bitset>#include <list>#include <sstream>#include <set>#include <functional>using namespace std;#define INT_MAX 1 << 30#define MAX 100typedef long long ll;int n,d;int x[10001],y[10001];char s;int p,q,p1,q1;stack<int> a,b;int par[10001],ran[10001];void init(int n){    for (int i = 1; i <= n; i += 1){        par[i] = i;        ran[i] = 0;    }}int root(int x){    if (par[x] == x){        return x;    }else{        return par[x] = root(par[x]);    }}void unite(int x,int y){    x = root(x);    y = root(y);    if (x == y) return;    if (ran[x] < ran[y]){        par[x] = y;    }else{        par[y] = x;        if (par[x] == par[y]){            ran[x]++;        }    }}bool same(int x,int y){    return root(x) == root(y);}int main(int argc, char const* argv[]){    scanf("%d%d",&n,&d);    for (int i = 0; i < n; i += 1){        scanf("%d%d",&x[i],&y[i]);    }    init(n);    while (scanf("%c",&s) != EOF){        //每次检测        if (s == 'O'){            scanf("%d",&p);            //遍历元素更新并查集            b = a;            while (!b.empty()){                q = b.top();                b.pop();                if (pow(double(x[p-1]-x[q-1]),2)+pow(double(y[p-1]-y[q-1]),2) <= pow(double(d),2)){                    unite(p,q);                }            }            a.push(p);        }else if(s == 'S'){            scanf("%d%d",&p1,&q1);            if (same(p1,q1))                printf("SUCCESS\n");            else                printf("FAIL\n");        }    }    return 0;}