【JZOJ5324】【GDOI2017模擬8.21】麻將堆

Description

Data Constraint

Solution

看到這種題就知道是凸包啦。
我們分別將第一層和第二層的麻將按x軸排序。枚舉第二層麻將，動態將麻將的y座標加入set，每次在set內查詢滿足y座標的點，並將相交的區域的四個頂點加入數組。可以證明最後與一個矩形相交的矩形的頂點最多隻有16個。我們將16個點按極角排序後做一遍凸包，判斷第二層的重心是否在凸包內即可。時間複雜度O(NlogN)。

Code

#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<set>
#define db double
using namespace std;
const int maxn=3e5+5;
const db f[4][2]={{1,1},{1,-1},{-1,-1},{-1,1}};
struct code{
    db x,y,z;int id;
}a[maxn],b[maxn],c[maxn];
struct code1{
    db x;int id;
    bool friend operator < (code1 x,code1 y){
        return x.x<y.x||x.x==y.x && x.id<y.id;
    }
}g;
int n,i,t,j,k,l,num,m,bz[maxn],bz1[4],d[maxn],num1,tot;
db p,q,pp,qq,x,y,z,xx,yy,zz,x2,y2;
set<code1>f1;
bool cmp(code x,code y){
    return x.z<y.z;
}
bool cmp1(code x,code y){
    return x.x<y.x;
}
db pan(int x,int y,int z){
    db xx=c[y].x-c[x].x,yy=c[y].y-c[x].y,x2=c[z].x-c[x].x,y2=c[z].y-c[x].y;
    return xx*y2-x2*yy;
}
void make(int j){
    int k,l;
    for (k=0;k<4;k++){
            xx=a[j].x+pp*f[k][0];yy=a[j].y+qq*f[k][1];
            if (xx<b[i].x-pp || xx>b[i].x+pp || yy<b[i].y-qq || yy>b[i].y+qq)continue;
            c[++num1].x=xx;c[num1].y=yy;
            x=xx-b[i].x;y=yy-b[i].y;
            c[num1].z=atan2(y,x);
            for (l=0;l<4;l++){
                x2=b[i].x+pp*f[l][0];y2=b[i].y+qq*f[l][1];
                if (x2<a[j].x-pp || x2>a[j].x+pp || y2<a[j].y-qq || y2>a[j].y+qq)continue;
                c[++num1].x=x2;c[num1].y=yy;
                x=x2-b[i].x;y=yy-b[i].y;
                c[num1].z=atan2(y,x);
                c[++num1].x=xx;c[num1].y=y2;
                x=xx-b[i].x;y=y2-b[i].y;
                c[num1].z=atan2(y,x);
                if (!bz1[l]){
                    c[++num1].x=xx;c[num1].y=yy;
                    x=xx-b[i].x;y=yy-b[i].y;
                    c[num1].z=atan2(y,x);
                }
            }
        }
}
int main(){
    freopen("heap.in","r",stdin);freopen("heap.out","w",stdout);
    scanf("%d%lf%lf",&n,&p,&q);pp=p/2;qq=q/2;
        for (i=1;i<=n;i++){
            scanf("%lf%lf%lf",&x,&y,&z);
            if (z==1) a[++num].x=x,a[num].y=y;
            else b[i-num].x=x,b[i-num].y=y,b[i-num].id=i,bz[i]=2;
        }
        sort(a+1,a+num+1,cmp1);
        m=n-num;sort(b+1,b+m+1,cmp1);tot=1;k=1;g.x=1e10;f1.insert(g);
        for (i=1;i<=m;i++){
            num1=0;memset(bz1,0,sizeof(bz1));
            while (a[tot].x-b[i].x<=p && tot<=num) g.x=a[tot].y+qq,g.id=tot++,f1.insert(g);
            while (b[i].x-a[k].x>p && k<=tot) g.x=a[k].y+qq,g.id=k++,f1.erase(g);
            x=b[i].y-qq;g.x=x-1;g.id=0;
            while (1){
                g=*f1.upper_bound(g);
                if (g.x-q>b[i].y+qq) break;
                make(g.id);
            }
            sort(c+1,c+num1+1,cmp);t=0;
            for (j=1;j<=num1;j++)
                if (c[j].x!=c[t].x || c[j].y!=c[t].y) c[++t]=c[j];
            num1=t;
            if (num1==1){
                if (c[1].x==b[i].x && c[1].y==b[i].y) bz[b[i].id]=1;
            }else if (num1==2){
                x=c[1].x-c[2].x;y=c[1].y-c[2].y;
                xx=c[1].x-b[i].x;yy=c[1].y-b[i].y;
                if (x*yy==xx*y) bz[b[i].id]=1;
            }else if (num1>2){
                d[1]=1;d[2]=2;d[0]=d[3]=3;
                for (j=4;j<=num1;j++){
                    while (pan(d[d[0]],d[d[0]-1],j)>0 && d[0]>1) d[0]--;
                    d[++d[0]]=j;
                }
                x=xx=d[1];c[0]=b[i];
                for (j=2;j<=d[0];j++){
                    if (pan(0,x,d[j])>=0 && pan(0,1,d[j])>=0) x=d[j];
                    if (pan(0,xx,d[j])<=0 && pan(0,1,d[j])<=0) xx=d[j];
                }
                if (pan(0,x,xx)>=0) bz[b[i].id]=1;
            }
        }t=0;
        for (i=1;i<=n;i++)
            if (bz[i]==2) t++;
        printf("%d\n",t);
        for (i=1;i<=n;i++)
            if (bz[i]==2) printf("%d ",i);
}