• BZOJ2429: [HAOI2006]聪明的猴子


    题目:http://www.lydsy.com/JudgeOnline/problem.php?id=2429

    题解:从某一点遍历n个点,且使最长边最短,就是MST了。

    代码:

      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,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)mx=d[x];
     81         for1(i,n)if(!v[i]&&dist(i,x)<d[i])
     82         {
     83            d[i]=dist(i,x);
     84            q.push(pa(d[i],i));
     85         }
     86     }
     87 }
     88 
     89 int main()
     90 
     91 {
     92 
     93     freopen("input.txt","r",stdin);
     94 
     95     freopen("output.txt","w",stdout);
     96 
     97     m=read();
     98     for1(i,m)b[i]=read();
     99     n=read();
    100     for1(i,n)a[i].x=read(),a[i].y=read();
    101     prim();
    102     int ans=0;
    103     for1(i,m)if(b[i]>=mx)ans++;
    104     cout<<ans<<endl;
    105 
    106     return 0;
    107 
    108 }  
    View Code
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  • 原文地址:https://www.cnblogs.com/zyfzyf/p/4255565.html
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