分配问题，将小精灵分配给巫妖，巫妖可接收量取决于时间长短，适应性取决于地形

预处理巫妖小精灵对的可行性，就是询问线段 \(AB\) 与圆 \(O\) 是否有交点

过点 \(O\) 作直线 \(AB\) 的垂线，考虑垂足，若垂足不在线段 \(AB\) 上，则问题等价于线段是否存在一段点在圆内

否则通过 \(\overrightarrow{AO} \ast \overrightarrow{AB}\) 计算 \(\overrightarrow{AO}\), \(\overrightarrow{AB}\) 所夹平行四边形面积，进而求得 \(O\) 到直线 \(AB\) 的距离，与半径比较大小即可

列出式子整理一下就可以避免double了！还有不要忘了巫妖有攻击半径

二分时间，用网络流判断可行性，即

由源点向巫妖连施法次数的边，巫妖向小精灵连边，小精灵连向汇点

二分上界为 \(MaxT\ast M\) ，下界为 \(0\) ，最大流等于小精灵数即为可行

#include <cstdio>
#include <algorithm>using std::min;
using std::max;const int MAXN=211;
const int MAXM=211;
const int MAXK=211;
const int MAXV=MAXN+MAXM;
const int MAXE=MAXN*MAXM+MAXN+MAXM;
const int INF=1034567890;int N, M, K;
bool Win=true;struct Pos{int x, y;Pos(){}Pos(int _x, int _y){x=_x;y=_y;}void read(){scanf("%d%d", &x, &y);}
};Pos operator - (Pos A, Pos B){return Pos(A.x-B.x, A.y-B.y);
}int Len(Pos A){return A.x*A.x+A.y*A.y;
}int Dis(Pos A, Pos B){return Len(A-B);
}int corss(Pos A, Pos B){return A.x*B.y-A.y*B.x;
}int dot(Pos A, Pos B){return A.x*B.x+A.y*B.y;
}struct Elf{Pos P;bool OutofTree;bool Link;
} El[MAXM];struct Lich{Pos P;int R, T;bool OutofTree;
} L[MAXN];struct Tree{Pos P;int R;
} T[MAXK];bool Map[MAXN][MAXM];struct Vert{int FE;int Bfn, Lev;
} V[MAXV];int Vcnt;
int Lp[MAXN], Ep[MAXM];
int Sour, Sink;struct Edge{int x, y, f, next, neg;
} E[MAXE<<1];int Ecnt;void addE(int a, int b, int c=1){++Ecnt;E[Ecnt].x=a;E[Ecnt].y=b;E[Ecnt].f=c;E[Ecnt].next=V[a].FE;V[a].FE=Ecnt;E[Ecnt].neg=Ecnt+1;++Ecnt;E[Ecnt].y=a;E[Ecnt].x=b;E[Ecnt].f=0;E[Ecnt].next=V[b].FE;V[b].FE=Ecnt;E[Ecnt].neg=Ecnt-1;
}int Q[MAXV], Head, Tail;
int BFN;bool BFS(int at=Sour){Head=Tail=1;++BFN;V[at].Lev=1;Q[Tail++]=at;V[at].Bfn=BFN;while(Head<Tail){at=Q[Head++];for(int k=V[at].FE, to;k>0;k=E[k].next){if(E[k].f<=0)   continue;to=E[k].y;if(V[to].Bfn==BFN)  continue;V[to].Lev=V[at].Lev+1;Q[Tail++]=to;V[to].Bfn=BFN;}}return V[Sink].Bfn==BFN;
}int DFS(int at=Sour, int inc=INF){if(at==Sink || inc<=0)  return inc;int ret=0, out;for(int k=V[at].FE, to;k>0;k=E[k].next){if(E[k].f<=0)   continue;to=E[k].y;if(V[to].Lev!=V[at].Lev+1)  continue;out=DFS(to, min(E[k].f, inc));ret+=out;inc-=out;E[k].f-=out;E[E[k].neg].f+=out;}if(inc>0)   V[at].Lev=-1;return ret;
}int DINIC(){int ret=0;while(BFS()){ret+=DFS();}return ret;
}int Left, Right, Mid;
int MaxT;int Cnt(Lich &l, int &t){if(l.T==0)  return INF;return t/(l.T)+1;
}bool Test(int Time){for(int i=1;i<=Vcnt;++i)    V[i].FE=0;Ecnt=0;for(int i=1;i<=N;++i)   addE(Sour, Lp[i], Cnt(L[i], Time));for(int i=1;i<=M;++i)   addE(Ep[i], Sink);for(int i=1;i<=N;++i)for(int j=1;j<=M;++j)if(Map[i][j])addE(Lp[i], Ep[/*i*/j]);return DINIC()==M;
}int main(){scanf("%d%d%d", &N, &M, &K);for(int i=1;i<=N;++i){L[i].P.read();scanf("%d%d", &L[i].R, &L[i].T);}for(int i=1;i<=M;++i)El[i].P.read();for(int i=1;i<=K;++i){T[i].P.read();scanf("%d", &T[i].R);}for(int i=1;i<=N;++i){L[i].OutofTree=true;for(int j=1;j<=K;++j)if(Dis(L[i].P, T[j].P)<=T[j].R*T[j].R)L[i].OutofTree=false;}for(int i=1;i<=M;++i){El[i].OutofTree=true;for(int j=1;j<=K;++j){if(Dis(El[i].P, T[j].P)<=T[j].R*T[j].R)El[i].OutofTree=false;}}for(int i=1;i<=M;++i)if(!El[i].OutofTree){Win=false;break;}if(!Win){puts("-1");return 0;}for(int i=1;i<=N;++i){if(!L[i].OutofTree) continue;for(int j=1, c;j<=M;++j){if(!El[j].OutofTree)    continue;if(Dis(El[j].P, L[i].P)>L[i].R*L[i].R)  continue;c=1;for(int k=1, Dot, Corss;k<=K && c>0;++k){Dot=dot(El[j].P-L[i].P, T[k].P-L[i].P);if(Dot>0 && Dot<Dis(L[i].P, El[j].P)){Corss=corss(El[j].P-L[i].P, T[k].P-L[i].P);if(1LL*Corss*Corss<=1LL*Dis(L[i].P, El[j].P)*T[k].R*T[k].R)c=0;}}if(c>0){//addE(Lp[i], Ep[j]);//printf("%d %d\n", i, j);Map[i][j]=true;El[j].Link=true;}}}for(int i=1;i<=M;++i)if(!El[i].Link){Win=false;break;}if(!Win){puts("-1");return 0;}for(int i=1;i<=N;++i)   Lp[i]=++Vcnt;for(int i=1;i<=M;++i)   Ep[i]=++Vcnt;Sour=++Vcnt;Sink=++Vcnt;for(int i=1;i<=N;++i)   MaxT=max(MaxT, L[i].T);Left=0;Right=MaxT*M;while(Left<Right){Mid=(Left+Right)>>1;if(Test(Mid))   Right=Mid;else    Left=Mid+1;}Mid=(Left+Right)>>1;printf("%d\n", Mid);return 0;
}/*
2 3 1
-100 0 100 3
100 0 100 5
-100 -10
100 10
110 11
5 5 105*/