poj Wireless Network(基础并查集)

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Wireless Network

题目链接：Click Here~

题目分析：

一道简单的并查集，但是在考虑如何加点的时候被卡了半天。kao！！！后来想用floyd处理连通的，一算O（10^14）啊！！！！！最后，才明白在加点的时候是把当前点（A）的与其他点（B）的距离dist(AB)<=d的加入，而且前提是B点必须是可行的！！就是这里卡了半天。后来发现题目是有提醒的，有一次被英语给坑了(-｡-;)。there is a computer C that can communicate with both A and B. 意思是通过可行的C连通A,B。开始给理解错了。别的就是基础了，细节看代码。

#include <iostream>#include <vector>#include <cstdio>#include <cstring>#include <cmath>using namespace std;const int maxn = 1005;struct Point{   double x,y;}p[maxn];vector<int> G[maxn];int n,f[maxn],cnnect[maxn];void Init(){    for(int i = 0;i < maxn;++i)     G[i].clear(),f[i] = i,cnnect[i] = 0;}double Distance(Point a,Point b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int Find(int x){    if(x==f[x])      return f[x];    return f[x] = Find(f[x]);}void Union(int u,int v){    int a = Find(u),b = Find(v);    if(a != b)       f[a] = b;}int main(){    double d;    while(~scanf("%d%lf",&n,&d))    {        Init();        for(int i = 1;i <= n;++i)           scanf("%lf%lf",&p[i].x,&p[i].y);        for(int i = 1;i <= n;++i)          for(int j = 1;j <= n;++j){             double dist = Distance(p[i],p[j]);             if(dist <= d)                G[i].push_back(j);          }        char str[5];        int x,y;        while(~scanf("%s",str)){            if(str[0]=='O'){               scanf("%d",&x);               cnnect[x] = 1;               for(int i = 0;i < (int)G[x].size();++i){                   y = G[x][i];                   if(cnnect[y])        //kao !!!!!!!!!!!                     Union(x,y);               }            }            else{               scanf("%d%d",&x,&y);               int a = Find(x),b = Find(y);               if(a == b)                 puts("SUCCESS");               else                 puts("FAIL");            }        }    }    return 0;}

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