JZOJ5401. 【NOIP2017提高A组模拟10.8】Star Way To Heaven

题目大意

要求一个最小的半径，
以每个点为圆心画一个圆，上下边界各向外延伸，使得上下边界联通。

题解

用最小生成树，
这里是完全图，只能用prim算法。
O(k2)
最后在最小生成树上面，
从上边界走到下边界的路径上面的最大值就是答案。

code

#include<queue>#include<cstdio>#include<iostream>#include<algorithm>#include <cstring>#include <string.h>#include <cmath>#include <math.h>#define ll long long#define N 6003#define db double#define P putchar#define G getchar#define mo 998244353using namespace std;char ch;void read(int &n){    n=0;    ch=G();    while((ch<'0' || ch>'9') && ch!='-')ch=G();    ll w=1;    if(ch=='-')w=-1,ch=G();    while('0'<=ch && ch<='9')n=(n<<3)+(n<<1)+ch-'0',ch=G();    n*=w;}db max(db a,db b){return a>b?a:b;}db min(db a,db b){return a<b?a:b;}ll abs(ll x){return x<0?-x:x;}ll sqr(ll x){return x*x;}void write(ll x){if(x>9) write(x/10);P(x%10+'0');}struct node{    int x,y;}a[N];int n,m,k,w[N],pos,x;int nxt[N],to[N],b[N],tot;bool bz[N];db ans,t,f[N],v[N];db get(int i,int j){    if(i==0)return m-a[j].y;    if(j>k)return a[i].y;    return sqrt(sqr(a[i].x-a[j].x)*1.00+sqr(a[i].y-a[j].y)*1.00);}void dfs(int x,int fa,db mx){    if(x>k)    {        ans=mx;        return;    }    for(int i=b[x];i;i=nxt[i])        dfs(to[i],x,max(mx,v[i]));}void ins(int x,int y,db z){    nxt[++tot]=b[x];    to[tot]=y;    v[tot]=z;    b[x]=tot;}int main(){    freopen("starway.in","r",stdin);    freopen("starway.out","w",stdout);    memset(bz,1,sizeof(bz));    read(n);read(m);read(k);    for(int i=1;i<=k;i++)        read(a[i].x),read(a[i].y),f[i]=get(0,i);    x=0;f[k+1]=m;    while(x!=k+1)    {        t=2147483647;        for(int i=1;i<=k+1;i++)            if(bz[i] && f[i]<t)t=f[i],pos=i;        bz[pos]=0;ins(w[pos],pos,t);x=pos;        for(int i=1;i<=k+1;i++)            if(bz[i])f[i]=min(get(x,i),f[i]),w[i]=x;    }    dfs(0,-1,0);    printf("%.8lf",ans/2);}