A - Wireless Network POJ - 2236（并查集）

最新推荐文章于 2021-10-25 19:11:05 发布

原创最新推荐文章于 2021-10-25 19:11:05 发布 · 260 阅读

0 ·

CC 4.0 BY-SA版权

程序设计专栏收录该内容

34 篇文章

订阅专栏

本文介绍了一种在地震破坏后修复无线网络的算法。重点在于如何通过修复计算机并测试其通信能力来逐步恢复网络功能。文章详细解释了输入输出格式，并提供了一个示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目的意思很坑，如果A和B都和C可以连接的话，则A与B可以直接连接

Sample Output

FAIL
SUCCESS

#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#define eps 1e-6
#define INF 0x3f3f3f3f

using namespace std;
struct point
{
    int x,y;
};
point p[1010];
bool rep[1010];//记录每个点是否被修好
int n,d,a,b;
char c;
int father[1010];

int Find(int x)
{

    while(x!=father[x])
        x=father[x];
    return x;
}

void Union(int a,int b)
{
    int fa=Find(a);
    int fb=Find(b);
    if(fa!=fb)
    {
        father[fa]=fb;
    }
}

void init()
{
    for(int i=1;i<1010;i++)
    {
        father[i]=i;
        rep[i]=false;
    }

}

double dist(point a,point b)
{
    double dis=1.0*sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    return dis;
}

int main()
{
    init();
    scanf("%d%d",&n,&d);
    for(int i=1;i<=n;i++)
        scanf("%d%d",&p[i].x,&p[i].y);
    while(scanf("%c",&c)!=EOF)
    {
        if(c=='O')
        {
            scanf("%d",&a);
            rep[a]=true;
            for(int i=1;i<=n;i++)
            {
                if(i==a) continue;
                else
                {
                    if(rep[i]&&dist(p[i],p[a])<=d)
                    //判断已知的两个点是不是可以连同，两点之间线段的距离必然小于折线
                        Union(i,a);
                }
            }
        }
        if(c=='S')
        {
            scanf("%d%d",&a,&b);
            if(Find(a)==Find(b)) puts("SUCCESS");
            else puts("FAIL");
        }
    }
    return 0;
}