题目链接
http://acm.hdu.edu.cn/showproblem.php?pid=6089
题解
这波强行维护搞得我很懵逼。。。
扫描线，只考虑每个点能走到左上方(不包括正上方，但包括正左方)的哪些点，然后旋转四次坐标系处理
所有询问和操作点按照先\(x\)后\(y\)坐标的顺序排序，然后枚举每一行，按\(y\)从小到大的顺序枚举这一行每个点
对于一个询问点找出前面最后一个操作点，那么要求的就是一个矩形减去一个区间内所有后缀最大值的和
然后这个东西可以用线段树直接维护，记录个区间最大值然后pushup的时候二分
操作点就相当于单点修改
时间复杂度\(O(n\log^2n)\).
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<utility>
#include<algorithm>
#define llong long long
#define pll pair<llong,llong>
#define mkpr make_pair
using namespace std;
const int N = 1e5;
struct SegmentTree
{
struct SgTNode
{
llong maxi,psum;
} sgt[(N<<2)+3];
void clear(int u,int le,int ri)
{
sgt[u].maxi = sgt[u].psum = 0;
if(le==ri) return;
int mid = (le+ri)>>1;
clear(u<<1,le,mid); clear(u<<1|1,mid+1,ri);
}
llong calc(int u,int le,int ri,llong x)
{
if(le==ri) {return sgt[u].maxi<x ? x : sgt[u].maxi;}
int mid = (le+ri)>>1;
if(sgt[u<<1|1].maxi<x) {return calc(u<<1,le,mid,x)+x*(ri-mid);}
else {return sgt[u].psum-sgt[u<<1|1].psum+calc(u<<1|1,mid+1,ri,x);} //Note here!
}
void pushup(int u,int le,int ri)
{
int mid = (le+ri)>>1;
sgt[u].maxi = max(sgt[u<<1].maxi,sgt[u<<1|1].maxi);
sgt[u].psum = calc(u<<1,le,mid,sgt[u<<1|1].maxi)+sgt[u<<1|1].psum;
}
void modify(int u,int le,int ri,int pos,llong x)
{
if(le==pos && ri==pos) {sgt[u].maxi = sgt[u].psum = x; return;}
int mid = (le+ri)>>1;
if(pos<=mid) {modify(u<<1,le,mid,pos,x);}
else {modify(u<<1|1,mid+1,ri,pos,x);}
pushup(u,le,ri);
}
pll query(int u,int le,int ri,int lb,int rb,llong x)
{
if(lb>rb) {return make_pair(0,0);}
if(le>=lb && ri<=rb) {return make_pair(max(sgt[u].maxi,x),calc(u,le,ri,x));}
int mid = (le+ri)>>1;
if(rb<=mid) {return query(u<<1,le,mid,lb,rb,x);}
else if(lb>mid) {return query(u<<1|1,mid+1,ri,lb,rb,x);}
else
{
pll retr = query(u<<1|1,mid+1,ri,lb,rb,x);
pll retl = query(u<<1,le,mid,lb,rb,retr.first);
return mkpr(retl.first,retl.second+retr.second);
}
}
} sgt;
struct Query
{
int x,y,id;
Query() {}
Query(int _x,int _y,int _id) {x = _x,y = _y,id = _id;}
bool operator <(const Query &arg) const
{
return x<arg.x || (x==arg.x && y<arg.y);
}
};
llong fans[N+3];
int mx[N+3];
namespace Solve
{
Query qr[(N<<1)+3];
int q;
int nx,ny;
void addquery(int x,int y,int id) {q++; qr[q] = Query(x,y,id);}
void solve()
{
sort(qr+1,qr+q+1); int j = 1; while(j<=q && qr[j].x==0) j++;
for(int i=1; i<=nx; i++)
{
int k = 0;
while(j<=q && qr[j].x==i)
{
if(qr[j].id==0)
{
sgt.modify(1,1,ny,qr[j].y,i);
k = qr[j].y; mx[qr[j].y] = i;
}
else
{
llong ans = (llong)i*(qr[j].y-1-k)-sgt.query(1,1,ny,k+1,qr[j].y-1,mx[qr[j].y]).second; //Note here!
fans[qr[j].id] += ans;
}
j++;
}
}
q = 0; sgt.clear(1,1,ny); for(int i=1; i<=ny; i++) mx[i] = 0ll;
}
}
Query a[N+3],b[N+3];
int nx,ny,m,q;
int main()
{
int T; scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d",&nx,&ny,&m,&q);
for(int i=1; i<=m; i++) scanf("%d%d",&a[i].x,&a[i].y);
for(int i=1; i<=q; i++) scanf("%d%d",&b[i].x,&b[i].y);
//1
Solve::nx = nx; Solve::ny = ny;
for(int i=1; i<=m; i++) Solve::addquery(a[i].x,a[i].y,0);
for(int i=1; i<=q; i++) Solve::addquery(b[i].x,b[i].y,i);
Solve::solve();
//2
Solve::nx = ny; Solve::ny = nx;
for(int i=1; i<=m; i++) Solve::addquery(a[i].y,nx+1-a[i].x,0);
for(int i=1; i<=q; i++) Solve::addquery(b[i].y,nx+1-b[i].x,i);
Solve::solve();
//3
Solve::nx = nx; Solve::ny = ny;
for(int i=1; i<=m; i++) Solve::addquery(nx+1-a[i].x,ny+1-a[i].y,0);
for(int i=1; i<=q; i++) Solve::addquery(nx+1-b[i].x,ny+1-b[i].y,i);
Solve::solve();
//4
Solve::nx = ny; Solve::ny = nx;
for(int i=1; i<=m; i++) Solve::addquery(ny+1-a[i].y,a[i].x,0);
for(int i=1; i<=q; i++) Solve::addquery(ny+1-b[i].y,b[i].x,i);
Solve::solve();
for(int i=1; i<=q; i++) printf("%lld\n",fans[i]+1);
memset(fans,0,sizeof(fans));
}
return 0;
}
