[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
,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;
}