2019-2020 ACM-ICPC Brazil Subregional Programming Contest Problem A Artwork (并查集)

Posted 2020-12-12 lr599909928

• 题意:有一个矩形,有(k)个警报器,警报器所在半径(r)内不能走,问是否能从左上角走到右下角.

• 题解:用并查集将所有相交的圆合并,那么不能走的情况如下图所示

所以最后查询判断一下即可.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <bitset>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  
  struct misaka{
  	ll x;
  	ll y;
  	ll r;
  }s[N];
  
  int m,n,k;
  int p[N];
  
  int find(int x){
  	if(p[x]!=x) p[x]=find(p[x]);
  	return p[x];
  }
  
  ll pow(ll x){
  	return x*x;
  }
  
  bool judge(misaka a,misaka b){
  	return pow(a.x-b.x)+pow(a.y-b.y)<=a.r*a.r+b.r*b.r+2*a.r*b.r;
  }
  
  bool check(misaka a,misaka b){
  	if(a.x>a.r && a.y+a.r<m)  return false;
  	if(b.x+b.r<n && b.y-b.r>0) return false;
  	return true;
  }
  
  int main() {
  	scanf("%d%d%d",&n,&m,&k);
  	for(int i=1;i<=k;++i){
  		scanf("%lld%lld%lld",&s[i].x,&s[i].y,&s[i].r);
  		p[i]=i;
  	}
  	for(int i=1;i<=k;++i){
  		for(int j=i+1;j<=k;++j){
  			if(judge(s[i],s[j])){
  				int x=find(i);
  				int y=find(j);
  				if(x!=y){
  					p[x]=y;
  				}
  			}
  		}
  	}
  	for(int i=1;i<=k;++i){
  		find(i);            //压缩路径
  	}
  	bool ok=true;
  	for(int i=1;i<=k;++i){
  		for(int j=1;j<=k;++j){
  			if(check(s[i],s[j]) && p[i]==p[j])
  			   ok=false;
  		}
  	}
  	if(ok) puts("S");
  	else puts("N");
  	
      return 0;
  }
  
  

  参考于:https://blog.csdn.net/a1214034447/article/details/102818357