【计算几何】【二分答案】【最大流】bzoj1822 [JSOI2010]Frozen Nova 冷冻波

用三角形面积什么的算算点到直线的距离之类……其实相切的情况是可行的……剩下的就跟某SDOI2015一样了。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
#define N 201
#define EPS 0.000001
#define INF 2147483647
struct Point{int x,y;}jl[N];
typedef Point Vector;
typedef double db;
Vector operator - (Point a,Point b){return (Vector){a.x-b.x,a.y-b.y};}
int Cross(Vector a,Vector b){return a.x*b.y-a.y*b.x;}
int sqr(int x){return x*x;}
db dis(Point a,Point b){return sqrt((db)(sqr(a.x-b.x)+sqr(a.y-b.y)));}
struct WY{Point p; int r,t;}wy[N];
struct SHU{Point p; int r;}shu[N];
int Abs(int x){return x<0?(-x):x;}
bool can[N][N];
int n,m,K;
bool Can_Attack(WY a,Point b)
{
	if((db)a.r-dis(a.p,b)<EPS) return 0;
	for(int i=1;i<=K;++i)
	  if((db)Abs(Cross(a.p-b,shu[i].p-b))/dis(a.p,b)-(db)shu[i].r<-EPS)
	    return 0;
	return 1;
}
bool Must_Be_Attacked(int x)
{
	for(int i=1;i<=n;++i)
	  if(can[i][x])
	    return 1;
	return 0;
}
queue<int>q;
int nn,T,S;
int v[2*N*(N+1)],en,first[N*2+3],next[2*N*(N+1)],cap[2*N*(N+1)];
int d[N*2+3],cur[N*2+3];
void AddEdge(int U,int V,int Cap)
{
	v[en]=V; cap[en]=Cap; next[en]=first[U]; first[U]=en++;
	v[en]=U; cap[en]=0; next[en]=first[V]; first[V]=en++;
}
bool bfs()
{
	memset(d,-1,sizeof(int)*(nn+1));
	d[S]=0; q.push(S);
	while(!q.empty())
	  {
	  	int U=q.front(); q.pop();
	  	for(int i=first[U];i!=-1;i=next[i])
	  	  if(cap[i]&&d[v[i]]==-1)
	  	    {
	  	      d[v[i]]=d[U]+1;
	  	      q.push(v[i]);
	  	    }
	  }
	return d[T]!=-1;
}
int dfs(int U,int a)
{
	if(U==T||(!a)) return a;
	int Flow=0,f;
	for(int i=first[U];i!=-1;i=next[i])
	  if(d[v[i]]==d[U]+1&&(f=dfs(v[i],min(a,cap[i]))))
	    {
	      cap[i]-=f;
	      cap[i^1]+=f;
	      Flow+=f;
	      a-=f;
	      if(!a)
	        break;
	    }
	if(!Flow) d[U]=-1;
	return Flow;
}
int MaxFlow()
{
	int Flow=0,tmp;
	while(bfs())
	  {
	  	memcpy(cur,first,sizeof(int)*(nn+1));
	  	while(tmp=dfs(S,INF)) Flow+=tmp;
	  }
	return Flow;
}
bool check(int Lim)
{
	memset(first,-1,sizeof(int)*(nn+1));
	en=0;
	for(int i=1;i<=n;++i) AddEdge(S,1+i,(!Lim)?1:(Lim/wy[i].t+1));
	for(int i=1;i<=m;++i) AddEdge(1+n+i,T,1);
	for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) if(can[i][j]) AddEdge(1+i,1+n+j,INF);
	return MaxFlow()==m?1:0;
}
int main()
{
	scanf("%d%d%d",&n,&m,&K);
	for(int i=1;i<=n;++i) scanf("%d%d%d%d",&wy[i].p.x,&wy[i].p.y,&wy[i].r,&wy[i].t);
	for(int i=1;i<=m;++i) scanf("%d%d",&jl[i].x,&jl[i].y);
	for(int i=1;i<=K;++i) scanf("%d%d%d",&shu[i].p.x,&shu[i].p.y,&shu[i].r);
	for(int i=1;i<=n;++i)
	  for(int j=1;j<=m;++j)
	    can[i][j]=Can_Attack(wy[i],jl[j]);
	for(int j=1;j<=m;++j) if(!Must_Be_Attacked(j)) {puts("-1"); return 0;}
	T=nn=n+m+2; S=1;
	int l=0,r=4000000;
	while(r>l)
	  {
	  	int mid=(l+r>>1);
	  	if(check(mid)) r=mid;
	  	else l=mid+1;
	  }
	printf("%d\n",l);
	return 0;
}