BZOJ2429: [HAOI2006]聪明的猴子

  1 #include<cstdio>
  2 
  3 #include<cstdlib>
  4 
  5 #include<cmath>
  6 
  7 #include<cstring>
  8 
  9 #include<algorithm>
 10 
 11 #include<iostream>
 12 
 13 #include<vector>
 14 
 15 #include<map>
 16 
 17 #include<set>
 18 
 19 #include<queue>
 20 
 21 #include<string>
 22 
 23 #define inf 1000000000
 24 
 25 #define maxn 1000+5
 26 
 27 #define maxm 200000+5
 28 
 29 #define eps 1e-10
 30 
 31 #define ll long long
 32 
 33 #define pa pair<int,int>
 34 
 35 #define for0(i,n) for(int i=0;i<=(n);i++)
 36 
 37 #define for1(i,n) for(int i=1;i<=(n);i++)
 38 
 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 40 
 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 42 
 43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)
 44 
 45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)
 46 
 47 #define mod 1000000007
 48 #define sqr(x) (x)*(x)
 49 
 50 using namespace std;
 51 
 52 inline int read()
 53 
 54 {
 55 
 56     int x=0,f=1;char ch=getchar();
 57 
 58     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
 59 
 60     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}
 61 
 62     return x*f;
 63 
 64 }
 65 int n,m;
 66 double mx,mxx,b[maxn],d[maxn];
 67 bool v[maxn];
 68 struct rec{int x,y;}a[maxn];
 69 priority_queue<pa,vector<pa>,greater<pa> >q;
 70 inline double dist(int x,int y){return sqrt(sqr(a[x].x-a[y].x)+sqr(a[x].y-a[y].y));}
 71 inline void prim()
 72 {
 73     for1(i,n)d[i]=inf;
 74     d[1]=0;
 75     q.push(pa(0,1));
 76     while(!q.empty())
 77     {
 78         int x=q.top().second;q.pop();
 79         if(v[x])continue;v[x]=1;
 80         if(d[x]>mx)mxx=mx,mx=d[x];
 81         else if(d[x]>mxx)mxx=d[x];
 82         for1(i,n)if(!v[i]&&dist(i,x)<d[i])
 83         {
 84            d[i]=dist(i,x);
 85            q.push(pa(d[i],i));
 86         }
 87     }
 88 }
 89 
 90 int main()
 91 
 92 {
 93 
 94     freopen("input.txt","r",stdin);
 95 
 96     freopen("output.txt","w",stdout);
 97 
 98     m=read();
 99     for1(i,m)b[i]=read();
100     n=read();
101     for1(i,n)a[i].x=read(),a[i].y=read();
102     prim();
103     int ans=0;
104     for1(i,m)if(b[i]>=mx)ans++;
105     cout<<ans<<endl;
106 
107     return 0;
108 
109 }