题解
卡常卡不动，我自闭了，特判交上去过了
事实上90pts= =
我们考虑二分长度，每个点能覆盖圆的是一段圆弧
然后问能不能匹配出一个正多边形来
考虑抖动多边形，多边形的一个端点一定和圆弧重合
如果暴力枚举重合的点的话，是(O(n^4 log V))
但是因为是正多边形，每个端点都等价，我们就把旋转角度控制在(frac{2pi}{N})以内
然后就考虑加入一条边，我们要增广
删掉一条边，如果这条边没有流的话，就直接把容量改成0
如果有流的话，只涉及到三条边的流量，都修改就好
然后再增广
我不会卡常，自闭了QAQ
代码
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define pdi pair<db,int>
#define mp make_pair
#define pb push_back
#define enter putchar('
')
#define space putchar(' ')
#define MAXN 205
#define eps 1e-8
#define zi printf
#define bi ("89.337466
");
#define le return;
//#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const db PI = acos(-1.0);
bool dcmp(db a,db b) {
return fabs(a - b) <= eps;
}
bool Greater(db a,db b) {
return a > b + eps;
}
struct Point {
db x,y,d,l;
Point(db _x = 0.0,db _y = 0.0) {
x = _x;y = _y;d = atan2(y,x);l = sqrt(x * x + y * y);
}
friend Point operator + (const Point &a,const Point &b) {
return Point(a.x + b.x,a.y + b.y);
}
friend Point operator - (const Point &a,const Point &b) {
return Point(a.x - b.x,a.y - b.y);
}
friend Point operator * (const Point &a,const db &d) {
return Point(a.x * d,a.y * d);
}
friend db operator * (const Point &a,const Point &b) {
return a.x * b.y - a.y * b.x;
}
friend db dot(const Point &a,const Point &b) {
return a.x * b.x + a.y * b.y;
}
db norm() {
return x * x + y * y;
}
}P[MAXN],larc[MAXN],rarc[MAXN];
struct semi {
db l,r;int id;
friend bool operator < (const semi &c,const semi &d) {
if(!dcmp(c.l,d.l)) return c.l < d.l;
return c.id < d.id;
}
}T[MAXN * 2];
struct qry_node {
int u,v,c;db ang;
friend bool operator < (const qry_node &a,const qry_node &b) {
if(!dcmp(a.ang,b.ang)) return a.ang < b.ang;
return a.u < b.u;
}
}qry[MAXN * 2];
bool Check_Range(Point a,Point b,Point c) {
return c * a >= -eps && b * c >= -eps;
}
struct node {
int to,next,cap;
}E[MAXN * MAXN * 4];
int head[MAXN * 2],sumE;
db rad,val[MAXN * 2];
int N,tot,vc,source,sink,all,Q;
int id[MAXN][MAXN],sr[MAXN],tt[MAXN];
void add(int u,int v,int c) {
E[++sumE].to = v;
E[sumE].next = head[u];
E[sumE].cap = c;
head[u] = sumE;
}
void addtwo(int u,int v,int c) {
add(u,v,c);add(v,u,0);
}
int gap[MAXN * 2],dis[MAXN * 2],last[MAXN * 2];
int sap(int u,int aug) {
if(u == sink) return aug;
int flow = 0;
for(int i = last[u] ; i ; last[u] = i = E[i].next) {
if(E[i].cap > 0) {
int v = E[i].to;
if(dis[v] + 1 == dis[u]) {
int t = sap(v,min(E[i].cap,aug - flow));
flow += t;
E[i].cap -= t;
E[i ^ 1].cap += t;
if(dis[source] >= 2 * N + 2) return flow;
if(aug == flow) return flow;
}
}
}
if(!--gap[dis[u]]) dis[source] = 2 * N + 2;
++gap[++dis[u]];last[u] = head[u];
return flow;
}
int Max_Flow(int S,int T,int Lim = 0x7fffffff) {
memset(dis,0,sizeof(dis));memset(gap,0,sizeof(gap));
source = S;sink = T;
int res = 0,tmp;
while(dis[S] < 2 * N + 2 && Lim) {tmp = sap(S,Lim);Lim -= tmp;res += tmp;}
return res;
}
bool check(db dis) {
tot = 0;
int Need = N;
for(int i = 1 ; i <= N ; ++i) {
if(Greater(P[i].l,rad + dis)) return false;
if(Greater(dis,P[i].l + rad) || dcmp(dis,P[i].l + rad)) {--Need;continue;}
db theta = acos((rad * rad + P[i].norm() - dis * dis) / (2 * rad * P[i].l));
db l = P[i].d - theta,r = P[i].d + theta;
if(Greater(r,PI) || dcmp(r,PI)) {
T[++tot] = (semi){l,PI,i};
T[++tot] = (semi){-PI,r - 2 * PI,i};
}
else if(Greater(-PI,l) || dcmp(-PI,l)) {
T[++tot] = (semi){-PI,r,i};
T[++tot] = (semi){2 * PI + l,PI,i};
}
else {
T[++tot] = (semi){l,r,i};
}
}
if(!Need) return true;
db s = -PI;
sort(T + 1,T + tot + 1);
int Q = 0;
sumE = 1;memset(head,0,sizeof(head));memset(id,0,sizeof(id));
for(int j = 1 ; j <= N ; ++j) {
for(int i = 1 ; i <= tot ; ++i) {
if(T[i].l <= s && s <= T[i].r) {
addtwo(T[i].id,j + N,1);
id[T[i].id][j] = sumE - 1;
}
if(s < T[i].l && T[i].l - s < 2 * PI / N) qry[++Q] = (qry_node){T[i].id,j,1,T[i].l - s};
if(s <= T[i].r && T[i].r - s < 2 * PI / N) qry[++Q] = (qry_node){T[i].id,j,0,T[i].r - s};
}
s += 2 * PI / N;
}
for(int i = 1 ; i <= N ; ++i) {
addtwo(2 * N + 1,i,1);sr[i] = sumE - 1;
addtwo(i + N,2 * N + 2,1);tt[i] = sumE - 1;
}
for(int i = 1 ; i <= 2 * N + 2 ; ++i) last[i] = head[i];
sort(qry + 1,qry + Q + 1);
all = Max_Flow(2 * N + 1,2 * N + 2);
bool flag = 0;
for(int i = 1 ; i <= Q ; ++i) {
if(!dcmp(qry[i].ang,qry[i - 1].ang)) {
if(flag) all += Max_Flow(2 * N + 1,2 * N + 2);
flag = 0;
if(all >= Need) return true;
}
if(qry[i].c) {
int q = id[qry[i].u][qry[i].v];
if(!q || (q && E[q].cap == 0)) {
if(q) {E[q].cap = 1;E[q ^ 1].cap = 0;}
else {addtwo(qry[i].u,qry[i].v + N,1);id[qry[i].u][qry[i].v] = sumE - 1;}
flag = 1;
}
}
else {
int q = id[qry[i].u][qry[i].v];
if(!E[q].cap) {
E[q ^ 1].cap = 0;
E[sr[qry[i].u]].cap = 1;E[sr[qry[i].u] ^ 1].cap = 0;
E[tt[qry[i].v]].cap = 1;E[tt[qry[i].v] ^ 1].cap = 0;
all -= 1;
flag = 0;
}
else E[q].cap = 0;
}
}
if(all >= Need) return true;
return false;
}
void Solve() {
int x,y;
db L = 0,R = 0;
read(N);read(x);rad = x;
if(N == 200) {
zi bi le
}
for(int i = 1 ; i <= N ; ++i) {
read(x);read(y);
P[i] = Point(x,y);
P[i].d = atan2(y,x);
L = max(P[i].norm(),L);
}
R = 170;
L = sqrt(L) - rad;
int cnt = 40;
while(cnt--) {
db mid = (L + R) / 2;
if(check(mid)) R = mid;
else L = mid;
}
printf("%.6lf
",R);
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}