Ted has a new house

F - F
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d
& %I64u
Submit Status Practice HDU
3265

Description

Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. 

However, Ted is such a picky guy that in every poster he finds something ugly. So before he pastes a poster on the window, he cuts a rectangular hole on that poster to remove the ugly part. Ted is also a careless guy so that some of the pasted posters may overlap
when he pastes them on the window. 

Ted wants to know the total area of the window covered by posters. Now it is your job to figure it out. 

To make your job easier, we assume that the window is a rectangle located in a rectangular coordinate system. The window’s bottom-left corner is at position (0, 0) and top-right corner is at position (50000, 50000). The edges of the window, the edges of the
posters and the edges of the holes on the posters are all parallel with the coordinate axes. 

 

Input

The input contains several test cases. For each test case, the first line contains a single integer N (0<N<=50000), representing the total number of posters. Each of the following N lines contains 8 integers x1, y1, x2, y2, x3, y3,
x4, y4, showing details about one poster. (x1, y1) is the coordinates of the poster’s bottom-left corner, and (x2, y2) is the coordinates of the poster’s top-right corner. (x3, y3) is the coordinates of the hole’s bottom-left corner, while (x4, y4) is the
coordinates of the hole’s top-right corner. It is guaranteed that 0<=xi, yi<=50000(i=1…4) and x1<=x3<x4<=x2, y1<=y3<y4<=y2. 

The input ends with a line of single zero. 

 

Output

For each test case, output a single line with the total area of window covered by posters.
 

Sample Input

2
0 0 10 10 1 1 9 9
2 2 8 8 3 3 7 7
0

 

Sample Output

56

 

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cstdio>
using namespace std;
#define T 55555
typedef long long ll;
struct node
{
ll sum,flag;
}tr[T<<3];
struct rec
{
int x1,x2,y,flag;
rec(int a,int b,int c,int d){
x1=a,x2=b,y=c,flag=d;
}
bool operator<(const rec& a)const{
return y<a.y;
}
};
void pushup(int n,int ln,int rn)
{
if(tr
.flag)tr
.sum=rn-ln+1;
else if(ln==rn) tr
.sum=0;
else tr
.sum=tr[n<<1].sum+tr[n<<1|1].sum;
}
void update(int x1,int x2,int flag,int ln,int rn,int n)
{
if(x1<=ln&&x2>=rn){
tr
.flag+=flag;
pushup(n,ln,rn);
return;
}
int m=(ln+rn)>>1;
if(x1<=m)update(x1,x2,flag,ln,m,n<<1);
if(x2>m)update(x1,x2,flag,m+1,rn,n<<1|1);
pushup(n,ln,rn);
}
int main()
{
/*freopen("input.txt","r",stdin);*/
int n,x1,x2,x3,x4,y1,y2,y3,y4;
while(~scanf("%d",&n),n)
{
vector<rec> v;
long long sum=0;
while(n--)
{
scanf("%d%d%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3,&x4,&y4);
if(x1<x3)//拆成四个矩形
{
v.push_back(rec(x1,x3,y1,1));//底边给这段涂上颜色，
v.push_back(rec(x1,x3,y2,-1));//顶边把颜色去掉
}
if(x4<x2)
{
v.push_back(rec(x4,x2,y1,1));
v.push_back(rec(x4,x2,y2,-1));
}
if(y1<y3)
{
v.push_back(rec(x3,x4,y1,1));
v.push_back(rec(x3,x4,y3,-1));
}
if(y4<y2)
{
v.push_back(rec(x3,x4,y4,1));
v.push_back(rec(x3,x4,y2,-1));
}
}
sort(v.begin(),v.end());
memset(tr,0,sizeof(tr));
int end = v.size();
for(int i=0;i<end-1;++i){
if(v[i].x2>v[i].x1)
update(v[i].x1,v[i].x2-1,v[i].flag,0,T,1);
sum+=tr[1].sum*(v[i+1].y-v[i].y);
}
printf("%lld\n",sum);
}
return 0;
}