扫描线+线段树=矩形周长并
和求面积并差不多，线段树节点维护覆盖长度，线段分段数，是否覆盖，左右端点是否覆盖（判断线段分段数用 right.num+left.num-left.rb*right.lb）。一次扫描直接求出周长。。。（当然也可以求一遍x边长，再求一遍y边长。。。。这样简单的多）
线段排序时要注意Y相同时让左边的边在前，这样就解决重边问题了
Picture
| Time Limit: 2000MS | Memory Limit: 10000K | |
| Total Submissions: 9838 | Accepted: 5205 |
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.
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.
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Left(rt) rt<<1
#define Right(rt) rt<<1|1
const int maxn=5500;
struct Node
{
int l,r,len,num,cover,rb,lb;
}node[maxn<<2];
struct Line
{
int yl,yr,x,s;
bool operator < (const Line &M) const
{
if(x!=M.x) return x<M.x;
else return s>M.s;
}
}line[maxn<<1];
int Y[maxn<<1];
int find(int x,int k)
{
int low=1,high=k,mid;
while(low<=high)
{
mid=(low+high)/2;
if(Y[mid]==x) return mid;
else if(Y[mid]>x) high=mid-1;
else low=mid+1;
}
}
void push_up(int l,int r,int rt)
{
if(node[rt].cover)
{
node[rt].num=node[rt].rb=node[rt].lb=1;
node[rt].len=Y[r+1]-Y[l];
return ;
}
if(r==l)
{
node[rt].len=node[rt].num=node[rt].rb=node[rt].lb=0;
return ;
}
node[rt].len=node[Left(rt)].len+node[Right(rt)].len;
node[rt].num=node[Left(rt)].num+node[Right(rt)].num-node[Left(rt)].rb*node[Right(rt)].lb;
node[rt].rb=node[Right(rt)].rb;node[rt].lb=node[Left(rt)].lb;
}
void update(int l,int r,int rt,int L,int R,int c)
{
if(L<=l&&r<=R)
{
node[rt].cover+=c;
push_up(l,r,rt);
return ;
}
int m=(l+r)>>1;
if(L<=m) update(lson,L,R,c);
if(R>m) update(rson,L,R,c);
push_up(l,r,rt);
}
void build(int l,int r,int rt)
{
node[rt].l=l;
node[rt].r=r;
if(l==r) return ;
int m=(l+r)/2;
build(lson);build(rson);
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF&&n)
{
memset(node,0,sizeof(node));
memset(Y,0,sizeof(Y));
int x1,x2,y1,y2,cnt=0,k;
for(int i=0;i<n;i++)
{
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
line[++cnt]=(Line){y1,y2,x1,1};
Y[cnt]=y1;
line[++cnt]=(Line){y1,y2,x2,-1};
Y[cnt]=y2;
}
sort(line+1,line+1+cnt);
sort(Y+1,Y+1+cnt);
k=unique(Y+1,Y+1+cnt)-Y-1;
int ans=0,last=0;
for(int i=1;i<=cnt;i++)
{
int l=find(line[i].yl,k);
int r=find(line[i].yr,k)-1;
update(1,k-1,1,l,r,line[i].s);
if(i!=cnt) ans+=node[1].num*2*(line[i+1].x-line[i].x);//x
ans+=abs(node[1].len-last);//y
last=node[1].len;
}
printf("%d\n",ans);
}
return 0;
}