poj1177 Picture 扫描线求矩形周长并

Description

A number of rectangular posters, photographs and other pictures of the same shape are pasted on a wall. Their sides are all vertical or horizontal. Each rectangle can be partially or totally covered by the others. The length of the boundary of the union of
all rectangles is called the perimeter. 

Write a program to calculate the perimeter. An example with 7 rectangles is shown in Figure 1. 



The corresponding boundary is the whole set of line segments drawn in Figure 2. 



The vertices of all rectangles have integer coordinates. 

Input

Your program is to read from standard input. The first line contains the number of rectangles pasted on the wall. In each of the subsequent lines, one can find the integer coordinates of the lower left vertex and the upper right vertex of each rectangle. The
values of those coordinates are given as ordered pairs consisting of an x-coordinate followed by a y-coordinate. 

0 <= number of rectangles < 5000 

All coordinates are in the range [-10000,10000] and any existing rectangle has a positive area.

Output

Your program is to write to standard output. The output must contain a single line with a non-negative integer which corresponds to the perimeter for the input rectangles.

Sample Input

7
-15 0 5 10
-5 8 20 25
15 -4 24 14
0 -6 16 4
2 15 10 22
30 10 36 20
34 0 40 16

Sample Output

228

#include<iostream>
#include<cstring>
#include<cstdio>
#include<ostream>
#include<istream>
#include<algorithm>
#include<queue>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<vector>
#define fi first
#define se second
#define ll long long
#define pii pair<int,int>
#define inf (1<<30)
#define eps 1e-8
#define pb push_back
using namespace std;
const int maxn=10010;
struct Line
{
int x,y1,y2;
int flag;
}line[maxn];
int y[maxn];
struct Node
{
int cover,num,len;
bool lf,rf;
}node[maxn<<2];
int n;
bool cmp(const Line& a,const Line& b)
{
return a.x<b.x;
}
void build(int l,int r,int rt)
{
node[rt].cover=node[rt].num=node[rt].len=node[rt].lf=node[rt].rf=0;
if(l+1==r) return;
build(l,(l+r)>>1,rt<<1);
build((l+r)>>1,r,rt<<1|1);
}
void pushUp(int l,int r,int rt)
{
if(node[rt].cover) {
node[rt].num=1;
node[rt].len=y[r]-y[l];
node[rt].lf=node[rt].rf=1;
}
else if(l+1==r) {
node[rt].num=node[rt].len=node[rt].lf=node[rt].rf=0;
}
else {
node[rt].num=node[rt<<1].num+node[rt<<1|1].num-(node[rt<<1].rf&&node[rt<<1|1].lf);
node[rt].len=node[rt<<1].len+node[rt<<1|1].len;
node[rt].lf=node[rt<<1].lf;
node[rt].rf=node[rt<<1|1].rf;
}
}
void update(int L,int R,int flag,int l,int r,int rt)
{
if(L<=l&&R>=r) {
node[rt].cover+=flag;
pushUp(l,r,rt);
}
if(l+1==r) return;
int m=((l+r)>>1);
if(L<=m)
update(L,R,flag,l,m,rt<<1);
if(R>=m)
update(L,R,flag,m,r,rt<<1|1);
pushUp(l,r,rt);
}
int main()
{
int x1,x2,y1,y2;
while(~scanf("%d",&n)) {
for(int i=0;i<n;i++) {
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
line[i*2].x=x1;
line[i*2].y1=y1;
line[i*2].y2=y2;
line[i*2].flag=1;
line[i*2+1].x=x2;
line[i*2+1].y1=y1;
line[i*2+1].y2=y2;
line[i*2+1].flag=-1;
y[i*2]=y1; y[i*2+1]=y2;
}
sort(y,y+n*2);
int k=unique(y,y+n*2)-y;
sort(line,line+n*2,cmp);
build(0,k-1,1);
int ans=0;
int tot=n*2;
int pre=0;
for(int i=0;i<tot-1;i++) {
int L=lower_bound(y,y+k,line[i].y1)-y;
int R=lower_bound(y,y+k,line[i].y2)-y;
update(L,R,line[i].flag,0,k-1,1);
ans+=abs(node[1].len-pre)+node[1].num*2*(line[i+1].x-line[i].x);
pre=node[1].len;
}
ans+=pre;
printf("%d\n",ans);
}
return 0;
}