标签：dia div integer nbsp only and each ict else
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
这道题可以说是最简单的并查集，直接套模板其实就OK了
现附上AC代码：
#include<iostream>
#include<cstdio>
using namespace std;
struct node{
int a,plug;int x,y;
}com[1010]; //结构体变量存储a:并查集的标准量等于自身；plug:标志量，若为1则电脑修好，为0未修好；(x,y):存储电脑的坐标
int N,d;
int find(int x); //找到祖先
void unite(int i,int p); //合并并查集
int main(){
cin>>N>>d;
char w; //这个变量最扯了，刚开始没有用到这个变量，用的是：fflush(stdin)，结果wrong answer，改为%c,&w,吃掉空格换行符就可以，郁闷了。
for(int i=1;i<=N;i++)
scanf("%d%c%d%c",&com[i].x,&w,&com[i].y,&w),com[i].a=i,com[i].plug=0;//电脑坐标，以及初始化
char c;
int p,x,y;
//fflush(stdin);
while(~scanf("%c%c",&c,&w)){
if(c==‘O‘) {
scanf("%d%c",&p,&w);
for(int i=1;i<=N;i++)
if(com[i].plug&&(com[i].x-com[p].x)*(com[i].x-com[p].x)+(com[i].y-com[p].y)*(com[i].y-com[p].y)<=d*d) unite(i,p); //满足条件就合并
com[p].plug=1;
}
else {
scanf("%d%c%d%c",&x,&w,&y,&w);
if(find(x)==find(y)) printf("SUCCESS\n"); //查询是否可以相连
else printf("FAIL\n");
}
//fflush(stdin);
}
return 0;
}
int find(int x){
if(com[x].a==x) return x;
else return com[x].a=find(com[x].a);
}
void unite(int i,int p){
i=find(i),p=find(p);
com[i].a=p;
}
poj2236Wireless Network
标签：dia div integer nbsp only and each ict else
原文地址：https://www.cnblogs.com/sunjianzhao/p/11436818.html
