【bzoj1822】[JSOI2010] Frozen Nova 冷冻波

　　Circle的构造函数没赋初值都能过样例真是厉害啊……
　　这题做法比较显然。巫师和精灵构成了二分图，第i个巫师能干的精灵个数就是$\lfloor\frac{T_i}{time}+1\rfloor$，判断所有精灵能否被干完的话跑个最大流就完事了。
　　至于时间，显然具有单调性，于是二分之。
　　没了。
　　判断线段和圆相交的话另外写一篇文章好了。
　　设网络流时间复杂度为$O(F)$，则总时间复杂度为$O((F+nm)\log time)$。
　　网络流直接上了vani的模板。
　　

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=a,_=b;i<=_;i++)
#define per(i,a,b) for(int i=a,_=b;i>=_;i--)
#define For(i,a,b) for(int i=a,_=b;i< _;i++)
#define maxn 203

inline int rd() {
    char c = getchar();
    while (!isdigit(c) && c != '-') c = getchar() ; int x = 0 , f = 1;
    if (c == '-') f = -1 ; else x = c - '0';
    while (isdigit(c = getchar())) x = x * 10 + c - '0';
    return x * f;
}

//===========================NetworkFlow===============

const int NFmaxn = 3005;
const int NFmaxm = 600005;
const int NFinf = 0x7fffffff;

struct NF_Line
{
  int to, next, v, opt;
};

struct Network_Flow
{
  NF_Line li[NFmaxm];
  int be[NFmaxn], l, s, t, n, num[NFmaxn], note[NFmaxn];

  void makeline(int fr, int to, int v)
  {
    ++l;
    li[l].next = be[fr];
    be[fr] = l;
    li[l].to = to;
    li[l].v = v;
    li[l].opt = l + 1;

    ++l;
    li[l].next = be[to];
    be[to] = l;
    li[l].to = fr;
    li[l].v = 0;
    li[l].opt = l - 1;
  }

  void create(int N)
  {
    n = N;
  }

  int sap(int now, int maxf)
  {
    if (now == t) return maxf;
    int mi = n, tot = 0;
    for (int i = be[now]; i; i = li[i].next)
    {
      int to = li[i].to;
      if (li[i].v && note[to] + 1 == note[now])
      {
        int k = sap(to, min(maxf - tot, li[i].v));
        li[i].v -= k;
        tot += k;
        li[li[i].opt].v += k;
        if (note[s] >= n || tot == maxf) return tot;
      }
      if (li[i].v) mi = min(mi, note[to]);
    }
    if (!tot)
    {
      if (!--num[note[now]])
      {
        note[s] = n;
        return 0;
      }
      ++num[note[now] = mi + 1];
    }
    return tot;
  }

  int query(int S, int T)
  {
    s = S, t = T;
    memset(num, 0, sizeof(num));
    memset(note, 0, sizeof(note));
    num[0] = n;
    int ans = 0;
    while (note[s] < n) ans += sap(s, NFinf);
    return ans;
  }

  void clear()
  {
    l = s = t = n = 0;
    memset(be, 0, sizeof(be));
    memset(num, 0, sizeof(num));
    memset(note, 0, sizeof(note));
  }
}SAP;

//===========================NetworkFlow===============

vector<int> rel[maxn];

const int inf = 0x7fffffff;

inline void upmin(int&a , int b) { if (a > b) a = b ; }
inline void upmax(int&a , int b) { if (a < b) a = b ; }

template<class T> inline T sqr(T x) { return x * x ; }

struct Point {
    int x , y;
    Point(int x = 0 , int y = 0):x(x) , y(y) { }
    friend Point operator-(Point a , Point b) { return Point(a.x - b.x , a.y - b.y) ; }
    inline void print() { printf("%d %d\n" , x , y) ; }
}b[maxn];

struct Circle {
    Point o;
    int r;
    Circle() { o = Point() , r = 0 ; }
    Circle(Point o , int r):o(o) , r(r) { }
    inline void print() { printf("%d " , r) , o.print() ; }
}a[maxn] , c[maxn];

struct Line {
    Point a , b;
    Line() { a = b = Point();}
    Line(Point a , Point b):a(a) , b(b) {}
};

int n , m , K , mx , t[maxn];

void input() {
    n = rd() , m = rd() , K = rd();
    rep (i , 1 , n) {
        int x = rd() , y = rd() , r = rd();
        t[i] = rd();
        upmax(mx , t[i]);
        a[i] = Circle(Point(x , y) , r);
    }
    rep (i , 1 , m) {
        int x = rd() , y = rd();
        b[i] = Point(x , y);
    }
    rep (i , 1 , K) {
        int x = rd() , y = rd() , r = rd();
        c[i] = Circle(Point(x , y) , r);
    }
}

inline double cross(Point a , Point b) {
    return a.x * b.y - a.y * b.x;
}

inline double dot(Point a , Point b) {
    return a.x * b.x + a.y * b.y;
}

inline double DotToDotDis(Point a , Point b) {
    return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}

inline double DotToSegDis(Point p , Line l) {
    if (dot(l.a - p , l.b - p) > 0) return min(DotToDotDis(l.a , p) , DotToDotDis(l.b , p));
    return fabs(cross(p - l.a , l.b - l.a)) / DotToDotDis(l.a , l.b);
}

inline bool ok(Circle a , Point b) {
    if (DotToDotDis(a.o , b) > a.r) return 0;
    rep (k , 1 , K)
        if (DotToSegDis(c[k].o , Line(a.o , b)) < c[k].r) return 0;
    return 1;
}

void GetIntersect() {
    rep (i , 1 , n) rep (j , 1 , m) if (ok(a[i] , b[j]))
        rel[i].push_back(j);
}

int check(int mid) {
    SAP.clear();
    SAP.create(n + m + 2);
    rep (i , 1 , n) {
        SAP.makeline(n + m + 1 , i , mid / t[i] + 1);
        For (j , 0 , rel[i].size()) {
            int x = rel[i][j];
            SAP.makeline(i , x + n , 1);
        }
    }
    rep (i , 1 , m) SAP.makeline(i + n , n + m + 2 , 1);
    return SAP.query(n + m + 1 , n + m + 2);
}

void solve() {
    GetIntersect();
    int l = 0 , r = m * mx , ans = inf;
    while (l <= r) {
        int mid = (l + r) >> 1;
        int cnt = check(mid);
        if (cnt == m) upmin(ans , mid) , r = mid - 1;
        else l = mid + 1;
    }
    printf("%d\n" , ans == inf ? -1 : ans);
}

int main() {
    #ifndef ONLINE_JUDGE
        freopen("data.txt" , "r" , stdin);
    #endif
    input();
    solve();
    return 0;
}