传送门
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
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
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
给一堆点，两个操作查询是否连通和连通两个范围内的点。。。要在条件上细心一点，WA七发血的教训！！！
#include <iostream>
#include <cmath>
#include <string>
#include <cstring>
using namespace std;
typedef long long LL;
int pre[5000], point[5000][2];
bool fixed_computer[1200];
int Find(int x) { return (x == pre[x]) ? x : pre[x] = Find(pre[x]); }
void mix(int x, int y)
{
int fx = Find(x), fy = Find(y);
if (fx != fy)
{
pre[fy] = fx;
}
}
inline double dis(int x1, int y1, int x2, int y2)
{
return sqrt((x2 - x1)*(x2 - x1) + (y2 - y1)*(y2 - y1));
}
int main()
{
memset(fixed_computer, false, sizeof(fixed_computer));
int n, d;
cin >> n >> d;
for (int i = 1; i <= n; i++)
{
cin >> point[i][0] >> point[i][1];
pre[i] = i;//代表元为自己
}
string oper;
while (cin>>oper)
{
if (oper == "S")
{
int a, b;
cin >> a >> b;
(Find(a) == Find(b)) ? cout << "SUCCESS" << endl : cout << "FAIL" << endl;
}
else if (oper == "O")
{
int po;
cin >> po;
if (fixed_computer[po]) continue;
fixed_computer[po] = true;
for (int i = 1; i <= n; ++i)
{
if (fixed_computer[i] && dis(point[i][0],point[i][1],point[po][0],point[po][1])<=d)mix(i, po);
}
}
}
}