POJ-2236 Wireless Network 顺便讨论时间超限问题

Wireless Network

Time Limit: 10000MS		Memory Limit: 65536K
Total Submissions: 26131		Accepted: 10880

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

Source

POJ Monthly,HQM

从输入来讲题目的要求，首先输入n代表有n台破损的电脑，然后输入的是d，代表如果两台电脑之间的距离小于等于d则这两台电脑可以相连接。

然后输入O和一个数字代表维修这台电脑，输入S和两个数字a,b是判断a和b是否可以相连接（可以通过其他电脑来相连接），如可以则输出SUCCESS,否则输出FAIL

用并查集来做，如果两台电脑可以连接就用join加入一个集合（并查集概念），最后判断两个的pre就可以判断他们是否可以相连接

值得一提的是我写这题目的时候小小的探索了一下时间问题，发现用scanf确实比cin快（尽管在某些题目cin又比scanf快，好烦呀！！！）用int flag 比 bool flag 快（差不多...）

最后附一张探索过程的截图吧   嘿嘿

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    int x,y;
    int pre;
} p[];
int flag[];
int n,d;
int findn(int x)
{
    return x == p[x].pre ? x : findn(p[x].pre);//注意一定要这样写，如果改成普通的while(x!=p[x].pre)就会时间超限！！！以后可以这样简写！
}
void join(struct node p1,struct node p2)
{
    int fx=findn(p1.pre),fy=findn(p2.pre);
    if(fx!=fy)
        if((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y)<=d*d)//判断是否符合加入集合的条件
            p[fy].pre = fx;
}
int main()
{
    scanf("%d %d",&n,&d);
    for(int i=; i<=n; i++)
        p[i].pre = i;
    memset(flag,,sizeof(flag));
    for(int i=; i<=n; i++)
        scanf("%d %d",&p[i].x,&p[i].y);
    char c;
    while(scanf("\n%c",&c)!=EOF)//scanf里面的\n用来吸收上一个scanf产生的空行
    {
        if(c=='O')
        {
            int a;
            scanf("%d",&a);
            flag[a] = ;
            for(int i=; i<=n; i++)
            {
                if(flag[i]&&a!=i)//除i自己外其他点均与他建立关系，至于加不加入集合在join里面判断
                    join(p[i],p[a]);
            }
        }
        else
        {
            int a,b;
            scanf("%d %d",&a,&b);
            if(findn(a)==findn(b))
                printf("SUCCESS\n");
            else printf("FAIL\n");
        }
    }
    return ;
}