Grandpa Giuseppe won a professional pizza cutter, the kind of type reel and, to celebrate, baked a rectangle pizza to his grandchildren! He always sliced his pizzas into pieces by making cuts over continuous lines, not necessarily rectilinear, of two types: some begin at the left edge of the pizza, follow continuously to the right and end up in the right edge; other start on lower edge, follow continuously up and end up on the top edge. But Grandpa Giuseppe always followed a property: two cuts of the same type would never intersect. Here is an example with 4 cuts, two of each type, in the left part of the figure, which divide the pizza in 9 pieces.
It turns out that Grandpa Giuseppe simply loves geometry, topology, combinatorics and stuff; so, he decided to show to his grandchildren who could get more pieces with the same number of cuts if cross cuts of the same type were allowed. The right part of the figure shows, for example, that if the two cuts of the type that go from left to right could intercept, the pizza would be divided into 10 pieces.
Grandpa Giuseppe ruled out the property, but will not make random cuts. In addition to being one of the two types, they will comply with the following restrictions:
Two cuts have at most one intersection point and, if they have, it is because the cuts cross each other at that point;
Three cuts do not intersect in a single point;
Two cuts do not intersect at the border of the pizza;
A cut does not intercept a pizza corner.
Given the start and end points of each cut, your program should compute the number of resulting pieces from the cuts of Grandfather Giuseppe.
Input
The first line of the input contains two integers X and Y, (1≤X,Y≤109), representing the coordinates (X,Y) of the upper-right corner of the pizza. The lower left corner has always coordinates (0,0). The second line contains two integers H and V, (1≤H,V≤105), indicating, respectively, the number of cuts ranging from left to right and the number of cuts ranging from bottom to top. Each of the following lines H contains two integers Y1 and Y2, a cut that intercepts the left side with y-coordinate Y1 and the right side at y-coordinate Y2. Each of the following V lines contains two integers X1 and X2, a cut that intercept the bottom side at x-coordinate X1 and the upper side at x-coordinate X2.
Examples
Input
3 4
3 2
1 2
2 1
3 3
1 1
2 2
Output
13
Input
5 5
3 3
2 1
3 2
1 3
3 4
4 3
2 2
Output
19
Input
10000 10000
1 2
321 3455
10 2347
543 8765
Output
6
题意：
给你一个平面，n条横的线，n条竖的线，保证：
1。两条线最多只有一个交点
2.没有三条线交于一点
3.任意两条线不交于边界
4.任意线不经过角
问你这些线把这个平面分成几部分
题解：
切割平面的总和是横：1+与这条线相交的线的数量+纵：1+与这条线相交线的数量+横的线的数量*纵的线的数量+1
之后就可以用树状数组做了，一边排序，另一边加到树状数组里就可以得到有多少线与这个线相交。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=4e5+5;
struct cross//横
{
int lef,rig;
bool operator< (const cross& a)const
{
return lef<a.lef;
}
}cro[N];
struct roll
{
int top,down;
bool operator< (const roll& a)const
{
return top<a.top;
}
}rol[N];
ll heng[N],shu[N];
int lsc[N],lsr[N];
int lowbit(int x)
{
return x&(-x);
}
void across(int x)
{
for(int i=x;i;i-=lowbit(i))
heng[i]++;
}
void aroll(int x)
{
for(int i=x;i;i-=lowbit(i))
shu[i]++;
}
ll qcross(int x)
{
ll ans=0;
for(int i=x;i<N;i+=lowbit(i))
ans+=heng[i];
return ans;
}
ll qroll(int x)
{
ll ans=0;
for(int i=x;i<N;i+=lowbit(i))
ans+=shu[i];
return ans;
}
int main()
{
int c,r;
scanf("%d%d",&c,&r);
int n,m;
scanf("%d%d",&n,&m);
int ctr=0,ctc=0;
for(int i=1;i<=n;i++)
scanf("%d%d",&cro[i].lef,&cro[i].rig),lsc[++ctc]=cro[i].lef,lsc[++ctc]=cro[i].rig;
for(int i=1;i<=m;i++)
scanf("%d%d",&rol[i].down,&rol[i].top),lsr[++ctr]=rol[i].down,lsr[++ctr]=rol[i].top;
sort(lsc+1,lsc+1+ctc),sort(lsr+1,lsr+1+ctr);
int allc=unique(lsc+1,lsc+1+ctc)-lsc-1,allr=unique(lsr+1,lsr+1+ctr)-lsr-1;
for(int i=1;i<=n;i++)
{
cro[i].lef=lower_bound(lsc+1,lsc+1+allc,cro[i].lef)-lsc;
cro[i].rig=lower_bound(lsc+1,lsc+1+allc,cro[i].rig)-lsc;
}
sort(cro+1,cro+1+n);
for(int i=1;i<=m;i++)
{
rol[i].top=lower_bound(lsr+1,lsr+1+allr,rol[i].top)-lsr;
rol[i].down=lower_bound(lsr+1,lsr+1+allr,rol[i].down)-lsr;
}
sort(rol+1,rol+1+m);
ll ans=0;
for(int i=1;i<=n;i++)
{
ans+=1+qcross(cro[i].rig);
across(cro[i].rig);
}
for(int i=1;i<=m;i++)
{
ans+=1+qroll(rol[i].down);
aroll(rol[i].down);
}
ans+=1+(ll)n*m;
printf("%lld\n",ans);
return 0;
}