[分治 单调栈] BZOJ 4237 稻草人

x排序 考虑分治

左下角在左半边 右上角在右半边 

两边各维护一个单调栈 然后左边查询在右边二分

细节自行在代码中领会

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;

inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}

inline void read(int &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int N=200005;

struct abcd{
int x,y;
abcd(int x=0,int y=0):x(x),y(y) { }
}P
,s
,ql
,qr
;
int tl,tr;

inline bool cmpx(abcd A,abcd B){
return A.x<B.x;
}
inline bool cmpy(abcd A,abcd B){
return A.y>B.y;
}

int n; ll ans;
int tmp
;

void Solve(int l,int r){
if (l==r) return;
int mid=(l+r)>>1,x,y;
for (int i=l;i<=r;i++) s[i]=P[i];
sort(s+l,s+mid+1,cmpy);
sort(s+mid+1,s+r+1,cmpy);
tr=tl=0;
qr[0].y=ql[0].y=1<<30; int pnt=mid+1;
for (int i=l;i<=mid;i++)
{
while (pnt<=r && s[pnt].y>s[i].y)
{
while (tr && s[pnt].x<qr.x) tr--;
qr[++tr]=s[pnt]; pnt++;
}
while (tl && s[i].x>ql[tl].x) tl--;
ql[++tl]=s[i];
x=lower_bound(qr+1,qr+tr+1,ql[tl-1],cmpy)-qr;
y=lower_bound(qr+1,qr+tr+1,ql[tl],cmpy)-qr-1;
if (x<=y) ans+=y-x+1;
}
Solve(l,mid); Solve(mid+1,r);
}

int main()
{
int ix,iy;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n);
for (int i=1;i<=n;i++)
read(ix),read(iy),P[i]=abcd(ix,iy);
sort(P+1,P+n+1,cmpx);
Solve(1,n);
printf("%lld\n",ans);
return 0;
}