这个题呢就是一个线段树+容斥原理的题,二维线段树我也不会这个题就是用两个线段树,一颗线段树就是求有多少列被修改过,另一颗线段树就是求有多少行被修改过,每次修改就是将他所在的行和列都修改一下,当询问的时候就询问x1到x2的和与y1到y2的和然后进行容斥原理,就可以得到答案了,以下是代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
int l,r;
int sum;
}t1[100001<<6],t2[100001<<6];
void pushup1(int tr)
{
t1[tr].sum=t1[tr<<1].sum+t1[tr<<1|1].sum;
return;
}
void build1(int l,int r,int tr)
{
t1[tr].l=l;
t1[tr].r=r;
if(l==r)
{
t1[tr].sum=0;
return;
}
int mid=l+r>>1;
build1(l,mid,tr<<1);
build1(mid+1,r,tr<<1|1);
pushup1(tr);
return;
}
void change1(int l,int r,int tr)
{
if(t1[tr].l==t1[tr].r)
{
t1[tr].sum^=1;
return;
}
int mid=t1[tr].l+t1[tr].r>>1;
if(l<=mid)change1(l,r,tr<<1);
else change1(l,r,tr<<1|1);
pushup1(tr);
return;
}
int query1(int l,int r,int tr)
{
if(t1[tr].l>=l&&r>=t1[tr].r)
{
return t1[tr].sum;
}
int ret=0,mid=t1[tr].l+t1[tr].r>>1;
if(l<=mid)ret+=query1(l,r,tr<<1);
if(mid<r)ret+=query1(l,r,tr<<1|1);
return ret;
}
void pushup2(int tr)
{
t2[tr].sum=t2[tr<<1].sum+t2[tr<<1|1].sum;
return;
}
void build2(int l,int r,int tr)
{
t2[tr].l=l;
t2[tr].r=r;
if(l==r)
{
t2[tr].sum=0;
return;
}
int mid=l+r>>1;
build2(l,mid,tr<<1);
build2(mid+1,r,tr<<1|1);
pushup2(tr);
return;
}
void change2(int l,int r,int tr)
{
if(t2[tr].l==t2[tr].r)
{
t2[tr].sum^=1;
return;
}
int mid=t2[tr].l+t2[tr].r>>1;
if(l<=mid)change2(l,r,tr<<1);
else change2(l,r,tr<<1|1);
pushup2(tr);
return;
}
int query2(int l,int r,int tr)
{
if(t2[tr].l>=l&&r>=t2[tr].r)
{
return t2[tr].sum;
}
int ret=0,mid=t2[tr].l+t2[tr].r>>1;
if(l<=mid)ret+=query2(l,r,tr<<1);
if(mid<r)ret+=query2(l,r,tr<<1|1);
return ret;
}
int main()
{
long long n,m,q;
cin>>n>>m>>q;
build1(1,n,1);
build2(1,m,1);
for(int i=1;i<=q;i++)
{
int zb;
scanf("%d",&zb);
if(zb==1)
{
long long x,y;
cin>>x>>y;
//scanf("%d%d",&x,&y);
change1(x,x,1);
change2(y,y,1);
}
else
{
long long x1,x2,y1,y2;
cin>>x1>>y1>>x2>>y2;
//scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
long long a1,a2;
a1=query1(x1,x2,1);
a2=query2(y1,y2,1);
cout<<a1*(long long)(y2-y1+1)+a2*(long long)(x2-x1+1)-(long long)(a1*a2<<1)<<endl;
}
}
}
这个线段树看起来很易懂,求过