当前位置: 代码迷 >> 综合 >> POJ -2236 Wireless Network
  详细解决方案

POJ -2236 Wireless Network

热度:41   发布时间:2023-11-06 18:44:50.0

题意:

对于给出的电脑可以进行两项操作:修复和查询。

当两台电脑之间的距离小于d的时候就视为是连通的,套一下模板就可以。

今天又重新写了一次 发现得注意两点,距离不要sqrt会精度丢失,vis标记必须有。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
#define mem(a,x) memset(a,x,sizeof(a))
#define s1(x) scanf("%d",&x)
#define s2(x,y) scanf("%d%d",&x,&y)
#define s3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s4(x,y,z,k) scanf("%d%d%d%d",&x,&y,&z,&k)
#define ff(a,n) for(int i = 0 ; i < n; i++) scanf("%d",a+i)
#define tp(x) printf("x = %d\n",x)
#define ansp(x) printf("%d\n",x)
//inline ll ask(int x){ll res=0;while(x)res+=c[x],x-=x&(-x);return res;}
//inline void add(int x,int d){while(x<=n)c[x]+=d,x+=x&(-x);}
#define ls 2*rt
#define rs 2*rt+1
#define lson ls,L,mid
#define rson rs,mid+1,R
#define ll long long
using namespace std;
typedef pair<int,int> pii;
const ll inf = 0x3f3f3f3f;const int mx = 1e3+10;
int n,d,fa[mx];
bool vis[mx];
char cmd[10];
struct{int x,y;
}p[mx];
int find(int x){if(x!=fa[x])return fa[x] = find(fa[x]); return x;
}
void uni(int a, int b){int a1 = find(a);int b1 = find(b);int dos = (p[a].x-p[b].x)*(p[a].x-p[b].x)+(p[a].y-p[b].y)*(p[a].y-p[b].y);if(a1 != b1&& dos <= d*d)fa[a1] = b1;
}
int main(){//int T=10;	scanf("%d",&T);//	freopen("F:\\in.txt","r",stdin);s2(n,d);for(int i = 0; i < n ;i++){s2(p[i+1].x,p[i+1].y);}for(int i = 1; i <= n; i++)fa[i] = i;int a,b;while( scanf("%s",cmd) != EOF){if(cmd[0] == 'S'){s2(a,b);if(find(a) == find(b)){puts("SUCCESS");}elseputs("FAIL");}else{s1(a);vis[a] = 1;for(int i = 1; i <= n; i++){if(!vis[i] || i==a)continue;uni(a,i);}}}return 0;
}
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int fa[1100],vis[1100];
int n,d;
struct aa{int x,y;
}a[1100];
int find(int h)  
{  if (h!= fa[h])  {  fa[h] = find(fa[h]); //这个回溯时的压缩路径是精华  }  return fa[h];  
} 
/*int find(int x){while(x!=fa[x])     x=fa[x];return x; 
}*/
void uni(int c1,int c2)
{int dist=(a[c1].x-a[c2].x)*(a[c1].x-a[c2].x)+(a[c1].y-a[c2].y)*(a[c1].y-a[c2].y);c1=find(c1) ;c2=find(c2) ;if(dist<=d*d && c1!=c2){fa[c1]=c2 ;}
}
int main(){scanf("%d%d",&n,&d);for(int i=1;i<1100;i++)fa[i]=i;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++)scanf("%d%d",&a[i].x,&a[i].y);getchar();char te;while(cin>>te){if(te=='S'){int a1,b1;scanf("%d%d",&a1,&b1);if(find(a1)==find(b1)&&vis[a1]&&vis[b1]) puts("SUCCESS");else puts("FAIL");}else {int b1;scanf("%d",&b1);vis[b1]=1;for(int i=1;i<=n;i++){if(!vis[i]||i==b1) {continue;}uni(i,b1);}}} return 0;
}

 

 

 

 

 

  相关解决方案