Luogu-4048 [JSOI2010]冷冻波

Posted 2021-01-22 nianheng

考虑网络流，二分时间，源点向巫妖连流量为攻击次数的边，把每个巫妖向他能打的小精灵连一条流量为一的边，每个小精灵向汇点连一条边。

预处理每个小精灵能被那些巫妖打，这道题好像视线与树相切也算能打（雾。

#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e3+100,maxm=1e6+100,inf=0x7fffffff;
struct Point{
    double x,y;
    Point(double xx=0,double yy=0){
        x=xx,y=yy;
    }
}tre[maxn],a[maxn],b[maxn];
struct Vector{
    double x,y;
    Vector(double xx=0,double yy=0){
        x=xx,y=yy;
    }
};
struct Grafh{
    int v[maxm],w[maxm],nex[maxm],head[maxn],num;
    void add(int x,int y,int z){
        v[++num]=y;
        w[num]=z;
        nex[num]=head[x];
        head[x]=num;
        v[++num]=x;
        w[num]=0;
        nex[num]=head[y];
        head[y]=num;
    }
    void clean(){
        memset(head,0,sizeof(head));
        num=1;
    }
    Grafh(){num=1;}
}g,h;
int dcmp(double x){return fabs(x)<1e-9?0:(x>0?1:-1);}
Vector operator - (Point a,Point b){return Vector(a.x-b.x,a.y-b.y);}
double operator * (Vector a,Vector b){return a.x*b.y-a.y*b.x;}
double dot(Vector a,Vector b){return a.x*b.x+a.y*b.y;}
double len(Vector a){return sqrt(dot(a,a));}
double dists(Point p,Point a,Point b){
    Vector v1=b-a,v2=p-a,v3=p-b;
    if(dcmp(dot(v1,v2))<=0) return len(v2);
    else if(dcmp(dot(v1,v3))>=0) return len(v3);
    return fabs(v1*v2/len(v1));
}
int n,m,p;
int cc[maxn],s,t;
int cur[maxn],tim[maxn];
double ar[maxn],tr[maxn];
queue<int>q;
bool bfs(){
    memset(cc,0,sizeof(cc));
    cc[s]=1;
    q.push(s);
    while(!q.empty()){
        int x=q.front();
        q.pop();
        for(int i=h.head[x];i;i=h.nex[i])
            if(h.w[i]&&!cc[h.v[i]])
                cc[h.v[i]]=cc[x]+1,q.push(h.v[i]);
    }
    return cc[t];
}
int dfs(int x,int ll){
    if(x==t) return ll;
    for(int &i=cur[x];i;i=h.nex[i])
        if(h.w[i]&&cc[h.v[i]]==cc[x]+1){
            int pp=dfs(h.v[i],ll>h.w[i]?h.w[i]:ll);
            if(pp){
                h.w[i]-=pp;
                h.w[i^1]+=pp;
                return pp;
            }
        }
    return 0;
}
int dinic(){
    int maxl=0,ll;
    while(bfs()){
        memcpy(cur,h.head,sizeof(cur));
        while(ll=dfs(s,inf))
            maxl+=ll;
    }
    return maxl;
}
bool getdist(){
    s=0,t=n+m+1;
    for(int i=1;i<=m;i++){
        bool cz=0;
        for(int j=1;j<=n;j++)
            if(dcmp(len(a[j]-b[i])-ar[j])<=0){
                bool ok=1;
                for(int k=1;k<=p;k++)
                    if(dcmp(dists(tre[k],b[i],a[j])-tr[k])<0){
                        ok=0;
                        break;
                    }
                if(ok){
                    g.add(j,i+n,inf);
                    cz=1;
                }
            }
        if(!cz){
            printf("-1
");
            return 1;
        }
    }
    for(int i=1;i<=m;i++)
        g.add(i+n,t,1);
    return 0;
}
bool check(int x){
    h=g;
    for(int i=1;i<=n;i++)
        h.add(s,i,x/tim[i]+1);
    if(dinic()==m) return 1;
    return 0;
}
void work(){
    int l=0,r=4000001,mid,ans=0;
    while(l<r){
        mid=l+r>>1;
        if(check(mid))
            ans=mid,r=mid;
        else
            l=mid+1;
    }
    printf("%d
",ans);
}
int main(){
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++)
        scanf("%lf%lf%lf%d",&a[i].x,&a[i].y,&ar[i],&tim[i]);
    for(int i=1;i<=m;i++)
        scanf("%lf%lf",&b[i].x,&b[i].y);
    for(int i=1;i<=p;i++)
        scanf("%lf%lf%lf",&tre[i].x,&tre[i].y,&tr[i]);
    if(getdist()) return 0;
    work();
    return 0;
}