P3297 [SDOI2013]逃考

标签：abs   inf   lag   printf   span   amp   lin   mes   front   

题意

两个亲戚间的范围的分界线必定为两者连线的中垂线，因此我们用半平面交\(O(n^2\log n)\)求出每个人的范围，之后相邻的两个范围连边跑最短路即可。

注意特判\(n=0\)的情况。

code:

#include<bits/stdc++.h>
using namespace std;
const int maxn=610;
const double eps=1e-8;
const double inf=1e12;
const double Pi=acos(-1.0);
int T,n,m,tot,cnt_edge,st;
int head[maxn],dis[maxn];
double sx,sy,limx,limy;
bool vis[maxn];
struct edge{int to,nxt;}e[maxn*maxn];
inline void add_edge(int u,int v)
{
    e[++cnt_edge].nxt=head[u];
    head[u]=cnt_edge;
    e[cnt_edge].to=v;
}
struct Point
{
    double x,y;
    inline double len(){return sqrt(x*x+y*y);}
    Point operator+(const Point a)const{return (Point){x+a.x,y+a.y};}
    Point operator-(const Point a)const{return (Point){x-a.x,y-a.y};}
    Point operator*(const double k){return (Point){x*k,y*k};}
    Point operator/(const double k){return (Point){x/k,y/k};}
    double operator*(const Point a)const{return x*a.y-y*a.x;}
    double operator&(const Point a)const{return x*a.x+y*a.y;}
}p[maxn];
inline int dcmp(double x)
{
    if(fabs(x)<=eps)return 0;
    return x<0?-1:1;
} 
inline Point get(Point a,Point b){return b-a;}
inline Point turn(Point a,double theta){return (Point){-a.y,a.x};}
struct Line
{
    Point p,v;int id;double theta;
    bool operator<(const Line& a)const
    {
        return !dcmp(theta-a.theta)?dcmp(get(p,v)*get(p,a.v))<0:dcmp(theta-a.theta)<0;
    }
}line[maxn],q[maxn];
inline Point getpoint(Line l1,Line l2)
{
    Point p1=l1.p,v1=l1.v,p2=l2.p,v2=l2.v;
    v1=get(p1,v1),v2=get(p2,v2);
    Point u=get(p1,p2);
    return p2+v2*(u*v1)/(v1*v2);
}
inline bool check(Line a,Line b,Line c)
{
    Point p=getpoint(a,b);
    return dcmp(get(c.p,c.v)*get(c.p,p))<=0;
}
inline void solve(int id)
{
    for(int i=1;i<=tot;i++)line[i].theta=atan2(line[i].v.y-line[i].p.y,line[i].v.x-line[i].p.x);
    sort(line+1,line+tot+1);
    int cnt=0;line[0].theta=inf;
    for(int i=1;i<=tot;i++)if(line[i].theta!=line[i-1].theta)line[++cnt]=line[i];
    tot=cnt;
    int l,r;
    q[l=r=1]=line[1];q[++r]=line[2];
    for(int i=3;i<=tot;i++)
    {
        while(l<r&&check(q[r-1],q[r],line[i]))r--;
        while(l<r&&check(q[l],q[l+1],line[i]))l++;
        q[++r]=line[i];
    }
    while(l<r&&check(q[r-1],q[r],q[l]))r--;
    while(l<r&&check(q[l],q[l+1],q[r]))l++;
    for(int i=l;i<=r;i++)add_edge(id,q[i].id);
    bool flag=1;
    for(int i=l;i<=r;i++)if(dcmp(get(q[i].p,q[i].v)*get(q[i].p,(Point){sx,sy}))<=0)flag=0;
    if(flag)st=id;
}
inline Line get_line(int x,int y)
{
    Point z=(p[x]+p[y])/2.0;
    return (Line){z,z+turn(get(p[x],p[y]),Pi/2.0),y,0};
}
inline void work(int id)
{
    tot=0;
    Point p1=(Point){0,0},p2=(Point){limx,0},p3=(Point){limx,limy},p4=(Point){0,limy};
    line[++tot]=(Line){p1,p2,n+1,0};
    line[++tot]=(Line){p2,p3,n+1,0};
    line[++tot]=(Line){p3,p4,n+1,0};
    line[++tot]=(Line){p4,p1,n+1,0};
    for(int i=1;i<=n;i++)if(i!=id)line[++tot]=get_line(id,i);
    solve(id);
}
inline void spfa()
{
    memset(dis,0x3f,sizeof(dis));
    queue<int>q;
    q.push(st);dis[st]=0;vis[st]=1;
    while(!q.empty())
    {
        int x=q.front();q.pop();vis[x]=0;
        for(int i=head[x];i;i=e[i].nxt)
        {
            int y=e[i].to;
            if(dis[y]>dis[x]+1)
            {
                dis[y]=dis[x]+1;
                if(!vis[y])q.push(y),vis[y]=1;
            }
        }
    }
}
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    scanf("%d",&T);
    while(T--)
    {
        memset(head,0,sizeof(head));
        cnt_edge=st=0;
        scanf("%d",&n);
        scanf("%lf%lf%lf%lf",&limx,&limy,&sx,&sy);
        for(int i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
        if(!n){puts("0");continue;}
        for(int i=1;i<=n;i++)work(i);
        spfa();
        printf("%d\n",dis[n+1]);
    }
    return 0;
} 

P3297 [SDOI2013]逃考

标签：abs   inf   lag   printf   span   amp   lin   mes   front   

原文地址：https://www.cnblogs.com/nofind/p/12244717.html