Picture

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.

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

228

 求矩形周长并

ac代码

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#define N 25010
#define INF 0x3f3f3f3f
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
using namespace std;
struct s
{
double x1,x2,y;
int flag;
}a[N<<2];
double __hash[N<<2],sum[N<<2];
int col[N<<2],lb[N<<2],rb[N<<2],num[N<<2];
int cmp(s a,s b)
{
return a.y<b.y;
}
int bseach(double key,int n)
{
int l=1,r=n;
while(l<=r)
{
int mid=(l+r)>>1;
if(__hash[mid]==key)
return mid;
if(__hash[mid]<key)
l=mid+1;
else
r=mid-1;
}
return l;
}
void build(int l,int r,int tr)
{
}
void pushup(int tr,int l,int r)
{
if(col[tr])
{
sum[tr]=__hash[r+1]-__hash[l];
lb[tr]=rb[tr]=1;
num[tr]=2;
}
else
{
if(l==r)
{
sum[tr]=0;
lb[tr]=rb[tr]=0;
num[tr]=0;
}
else
{
sum[tr]=sum[tr<<1]+sum[tr<<1|1];
lb[tr]=lb[tr<<1];
rb[tr]=rb[tr<<1|1];
num[tr]=num[tr<<1]+num[tr<<1|1];
if(lb[tr<<1|1]&&rb[tr<<1])
num[tr]-=2;
}
}
}
void update(int L,int R,int l,int r,int tr,int flag)
{
if(L<=l&&r<=R)
{
col[tr]+=flag;
pushup(tr,l,r);
return;
}
int mid=(l+r)>>1;
if(L<=mid)
update(L,R,l,mid,tr<<1,flag);
if(R>mid)
update(L,R,mid+1,r,tr<<1|1,flag);
pushup(tr,l,r);
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
double x1,x2,y1,y2,lx=INF,rx=-INF;
int i,k=1;
for(i=1;i<=n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
a[k].x1=x1;
a[k].x2=x2;
a[k].y=y1;
a[k].flag=1;
__hash[k++]=x1;
a[k].x1=x1;
a[k].x2=x2;
a[k].y=y2;
a[k].flag=-1;
__hash[k++]=x2;
}
sort(a+1,a+k,cmp);
sort(__hash+1,__hash+k);
memset(sum,0,sizeof(sum));
memset(col,0,sizeof(col));
memset(lb,0,sizeof(lb));
memset(rb,0,sizeof(rb));
memset(num,0,sizeof(num));
double ans=0,last=0;
for(i=1;i<=k-1;i++)
{
int x,y;
x=bseach(a[i].x1,k-1);
y=bseach(a[i].x2,k-1)-1;
if(x<=y)
update(x,y,1,k-1,1,a[i].flag);
ans+=num[1]*(a[i+1].y-a[i].y);
ans+=fabs(sum[1]-last);
last=sum[1];
}
printf("%.lfn",ans);
}
}

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