[bzoj] 2716 天使玩偶 || CDQ分治

原题

已知n个点有天使玩偶，有m次操作：

操作1：想起来某个位置有一个天使玩偶

操作2：询问离当前点最近的天使玩偶的曼哈顿距离

---

显然的CDQ问题，三维分别为时间，x轴，y轴。

但是这道题的问题在于最近距离怎么维护。

曼哈顿距离定义为|x2-x1|+|y2-y1|，所以把绝对值展开后一共有四种情况：

\(x2-x1+y2-y1 => x2+y2-(x1+y1)
x1-x2+y2-y1 => -x2+y2+(x1-y1)
x2-x1+y1-y2 => x2-y2+(y1-x1)
x1-x2+y1-y2 => -x2-y2+(x1+y1)\)

所以把四种情况都讨论一遍，每次处理为该情况的点对即可。

#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 500010
using namespace std;
int n,m,op,ans[N],mxy,f[1000010];
struct hhh
{
    int x,y,tm,op,num;
    bool operator < (const hhh &b) const
	{
	    return tm<b.tm;
	}
}p[2*N],tmp[2*N];

int read()
{
    int ans=0,fu=1;
    char j=getchar();
    for (;j<'0' || j>'9';j=getchar()) if (j=='-') fu=-1;
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

inline void add(int x,int y)
{
    while (x<=mxy) f[x]=max(f[x],y),x+=x&-x;
}

inline int query(int x)
{
    int ans=0xc0c0c0c0;
    while (x) ans=max(ans,f[x]),x-=x&-x;
    return ans;
}

inline void clear(int x)
{
    while (x<=mxy) f[x]=0xc0c0c0c0,x+=x&-x;
}

void solve(int l,int r)
{
    if (l==r) return ;
    int mid=(l+r)>>1;
    solve(l,mid);
    solve(mid+1,r);
    int l1=l,l2=mid+1;
    for (int i=l;i<=r;i++)
    {
	if (l1<=mid && (l2>r || p[l1].x<p[l2].x)) tmp[i]=p[l1++];
	else tmp[i]=p[l2++];
    }

    for (int i=l;i<=r;i++)
    {
	p[i]=tmp[i];
	if (p[i].tm>mid && p[i].op) ans[p[i].num]=min(ans[p[i].num],p[i].x+p[i].y-query(p[i].y));
	if (p[i].tm<=mid && !p[i].op) add(p[i].y,p[i].x+p[i].y);
    }
    for (int i=l;i<=r;i++)
    {
	if (p[i].tm<=mid && !p[i].op)
	    clear(p[i].y);
    }

    for (int i=l;i<=r;i++)
    {
	if (p[i].tm>mid && p[i].op) ans[p[i].num]=min(ans[p[i].num],p[i].x-p[i].y-query(mxy-p[i].y));
	if (p[i].tm<=mid && !p[i].op) add(mxy-p[i].y,p[i].x-p[i].y);
    }
    for (int i=l;i<=r;i++)
    {
	if (p[i].tm<=mid && !p[i].op)
	    clear(mxy-p[i].y);
    }

    for (int i=r;i>=l;i--)
    {
	if (p[i].tm>mid && p[i].op) ans[p[i].num]=min(ans[p[i].num],-p[i].x+p[i].y-query(p[i].y));
	if (p[i].tm<=mid && !p[i].op) add(p[i].y,p[i].y-p[i].x);
    }
    for (int i=r;i>=l;i--)
    {
	if (p[i].tm<=mid && !p[i].op)
	    clear(p[i].y);
    }

    for (int i=r;i>=l;i--)
    {
	if (p[i].tm>mid && p[i].op) ans[p[i].num]=min(ans[p[i].num],-p[i].x-p[i].y-query(mxy-p[i].y));
	if (p[i].tm<=mid && !p[i].op) add(mxy-p[i].y,-p[i].x-p[i].y);
    }
    for (int i=r;i>=l;i--)
    {
	if (p[i].tm<=mid && !p[i].op)
	    clear(mxy-p[i].y);
    }
}

int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    n=read();
    m=read();
    for (int i=1;i<=n;i++)
    {
	p[i].x=read()+1;
	p[i].y=read()+1;
	p[i].tm=i;
	mxy=max(mxy,p[i].y);
    }
    for (int i=1;i<=m;i++)
    {
	op=read();
	p[n+i].x=read()+1;
	p[n+i].y=read()+1;
	p[n+i].tm=n+i;
	mxy=max(mxy,p[n+i].y);
	if (op==2)
	    p[n+i].num=++ans[0],p[n+i].op=1;
    }
    for (int i=1;i<=ans[0];i++) ans[i]=0x3f3f3f3f;
    mxy++;
    for (int i=0;i<=mxy;i++) f[i]=0xc0c0c0c0;
    sort(p+1,p+n+m+1);
    solve(1,n+m);
    for (int i=1;i<=ans[0];i++)
	printf("%d\n",ans[i]);
    return 0;
}