luogu4169 [Violet]天使玩偶/SJY摆棋子

题目描述

题解:

用cdq分治求一个点左下方最近的点的距离,然后坐标系旋转。

代码:

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 300050
const int inf = 0x7fffffff;
inline int rd()
{
    int f=1,c=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){c=10*c+ch-'0';ch=getchar();}
    return f*c;
}
int n,m,ans[N],cnt,maxx,maxy;
struct pnt
{
    int x,y,tm,wt,id;
}p[2*N],tmp[2*N];
struct Pair
{
    int x,id;
}px[2*N],py[2*N];
int tx[2*N],ty[2*N];
bool cmp(Pair a,Pair b)
{
    return a.x<b.x;
}
bool cmpt(pnt a,pnt b)
{
    if(a.tm!=b.tm)return a.tm<b.tm;
    if(a.x!=b.x)return a.x<b.x;
    return a.y<b.y;
}
void rev_x()
{
    for(int i=1;i<=n;i++)
        p[i].x = maxx-p[i].x;
    sort(p+1,p+1+n,cmpt);
    for(int i=1;i<maxx;i++)
    {
        if(maxx-i>i)
            swap(tx[i],tx[maxx-i]);
        tx[i]=-tx[i];
    }
}
void rev_y()
{
    for(int i=1;i<=n;i++)
        p[i].y = maxy-p[i].y;
    sort(p+1,p+1+n,cmpt);
    for(int i=1;i<maxy;i++)
    {
        if(maxy-i>i)
            swap(ty[i],ty[maxy-i]);
        ty[i]=-ty[i];
    }
}
struct BIT
{
    int f[2*N];
    void init(){for(int i=1;i<=maxy;i++)f[i]=-inf;}
    void up(int x,int d){while(x<2*N)f[x]=max(f[x],d),x+=(x&-x);}
    int down(int x)
    {
        int ret = -inf;
        while(x)ret=max(ret,f[x]),x-=(x&-x);
        return ret;
    }
    void clear(int x){while(x<2*N&&f[x]!=-inf)f[x]=-inf,x+=(x&-x);}
}tr;
void Sort(int l,int r)
{
    int mid = (l+r)>>1;
    int i = l,j = mid+1,k = l;
    while(i<=mid&&j<=r)
    {
        while(i<=mid&&p[i].x<=p[j].x)
        {
            tmp[k] = p[i];
            i++,k++;
        }
        while(j<=r&&p[i].x>p[j].x)
        {
            tmp[k] = p[j];
            j++,k++;
        }
    }
    while(i<=mid)
    {
        tmp[k] = p[i];
        i++,k++;
    }
    while(j<=r)
    {
        tmp[k] = p[j];
        j++,k++;
    }
    for(i=l;i<=r;i++)p[i]=tmp[i];
}
void cdq(int l,int r)
{
    if(l==r)return ;
    int mid = (l+r)>>1;
    cdq(l,mid),cdq(mid+1,r);
    Sort(l,mid),Sort(mid+1,r);
    int i = l,j = mid+1;
    while(j<=r)
    {
        while(i<=mid&&p[j].x>=p[i].x)
        {
            if(p[i].wt)tr.up(p[i].y,tx[p[i].x]+ty[p[i].y]);
            i++;
        }
        if(p[j].id)
        {
            int now = tr.down(p[j].y);
            if(now!=-inf)ans[p[j].id] = min(ans[p[j].id],tx[p[j].x]+ty[p[j].y]-now);
        }
        j++;
    }
    for(i=i-1;i>=l;i--)
        if(p[i].wt)tr.clear(p[i].y);
}
int main()
{
    n = rd(),m = rd();
    for(int i=1;i<=n;i++)
    {
        px[i].x = rd(),px[i].id = i;
        py[i].x = rd(),py[i].id = i;
        p[i].tm = 0,p[i].wt = 1,p[i].id = 0;
    }
    for(int opt,x,y,i=1;i<=m;i++)
    {
        opt = rd(),x = rd(),y = rd();
        n++;
        px[n].x = x,px[n].id = n;
        py[n].x = y,py[n].id = n;
        if(opt==1)
        {
            p[n].tm = i,p[n].wt = 1,p[n].id = 0;
        }else
        {
            cnt++;
            p[n].tm = i,p[n].wt = 0,p[n].id = cnt;
        }
    }
    memset(ans,0x3f,sizeof(ans));
    sort(px+1,px+1+n,cmp);
    sort(py+1,py+1+n,cmp);
    for(int las=-1,i=1;i<=n;i++)
    {
        if(las!=px[i].x)las=px[i].x,maxx++,tx[maxx]=las;
        p[px[i].id].x = maxx;
    }
    for(int las=-1,i=1;i<=n;i++)
    {
        if(las!=py[i].x)las=py[i].x,maxy++,ty[maxy]=las;
        p[py[i].id].y = maxy;
    }
    maxx++,maxy++;
    tr.init();
    sort(p+1,p+1+n,cmpt);
    cdq(1,n);
    rev_x();
    cdq(1,n);
    rev_y();
    cdq(1,n);
    rev_x();
    cdq(1,n);
    for(int i=1;i<=cnt;i++)
        printf("%d\n",ans[i]);
    return 0;
}

 

posted @ 2018-12-28 18:29  LiGuanlin  阅读(157)  评论(0编辑  收藏  举报