[二分答案+对偶图 || 三角剖分] BZOJ 3007 拯救小云公主 && BZOJ 4219 跑得比谁都快

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#define cl(x) memset(x,0,sizeof(x))
#define V G[p].v
using namespace std;

inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') c=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=3005;

struct Point{
int x,y;
void read(){
::read(x); ::read(y);
}
friend double Dist(Point A,Point B){
return sqrt((double)(A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
}
}P
,pos;

struct edge{
int u,v;
int next;
};

edge G[N*N];
int head
,inum;

inline void add(int u,int v,int p){
G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;
}

int n,S,T;
double dis

;

int Q
,l,r;
int ins
;

inline bool bfs(){
l=r=-1;
Q[++r]=S; ins[S]=1; cl(ins);
while (l<r){
int u=Q[++l];
for (int p=head[u];p;p=G[p].next)
if (!ins[V])
{
Q[++r]=V; ins[V]=1;
if (V==T) return 1;
}
}
return 0;
}

inline bool Check(double mid){
cl(head); inum=0;
for (int i=1;i<=n;i++)
{
if (P[i].x-1<mid || pos.y-P[i].y<mid)
add(S,i,++inum);
if (pos.x-P[i].x<mid || P[i].y-1<mid)
add(i,T,++inum);
}
for(int i=1;i<=n;i++)
for (int j=1;j<i;j++)
if (dis[i][j]<2*mid)
add(j,i,++inum),add(i,j,++inum);
return !bfs();
}

inline void Bin()
{
double L=0,R=min(pos.x-1,pos.y-1),MID;
while(R-L>1e-4)
if(Check(MID=(L+R)/2))
L=MID;
else
R=MID;
printf("%.2lf\n",(L+R)/2);
}

int main()
{
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); pos.read();
S=n+1; T=n+2;
for (int i=1;i<=n;i++) P[i].read();
for (int i=1;i<=n;i++)
for (int j=1;j<i;j++)
dis[j][i]=dis[i][j]=Dist(P[i],P[j]);
Bin();
return 0;
}