POJ 1177 Picture

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

Output

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 <stdio.h>
#include <string.h>
#include <algorithm>
#define N 10003
#define lson l,m,k<<1
#define rson m,r,k<<1|1
using namespace std;
struct segment
{
  int len,cover;//研究发现每次更新后树中的覆盖区间有变化值，可以累加起来、就可以得周长
};
struct line1
{
  int x,y1,y2;
  int flag;
  bool operator<(const line1 &L) const
  {
      if(x==L.x)//开始没这个、遇到 2 （0,0）(1,1)   (1,0)  (2 1)我的结果就错了，不过提交时居然A了，
        return flag>L.flag;
      return x<L.x;
  }
};
struct line2
{
    int y,x1,x2;
    int flag;
    bool operator<(const line2 &L) const
  {
      if(y==L.y)
        return flag>L.flag;
      return y<L.y;
  }
};
segment st[N<<2];
line1 L1[N];
line2 L2[N];
int rcx[N],rcy[N];
void up1(int &k,int &l,int &r)
{
    if(st[k].cover)
      st[k].len=rcy[r]-rcy[l];
    else if(l+1==r)
          st[k].len=0;
          else
          st[k].len=st[k<<1].len+st[k<<1|1].len;
}
void up2(int &k,int &l,int &r)
{
    if(st[k].cover)
      st[k].len=rcx[r]-rcx[l];
    else if(l+1==r)
          st[k].len=0;
          else
          st[k].len=st[k<<1].len+st[k<<1|1].len;
}
void build(int l,int r,int k)
{
    st[k].cover=st[k].len=0;
    if(l+1==r)
      return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
int flag;
void update1(int &y1,int &y2,int l,int r,int k)
{
    if(y1<=rcy[l]&&rcy[r]<=y2)
    {
        st[k].cover+=flag;
        up1(k,l,r);
        return ;
    }
    int m=(l+r)>>1;
    if(y1<rcy[m]) update1(y1,y2,lson);
    if(y2>rcy[m]) update1(y1,y2,rson);
    up1(k,l,r);
}
void update2(int &x1,int &x2,int l,int r,int k)
{
    if(x1<=rcx[l]&&rcx[r]<=x2)
    {
        st[k].cover+=flag;
        up2(k,l,r);
        return ;
    }
    int m=(l+r)>>1;
    if(x1<rcx[m]) update2(x1,x2,lson);
    if(x2>rcx[m]) update2(x1,x2,rson);
    up2(k,l,r);
}
int main()
{
    int n,nl,pl,len;
    int i,j,k;
    int x1,x2,y1,y2;
    while(scanf("%d",&n)!=EOF)
    {
       for(j=i=0;i<n;i++)
       {
           scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
           L1[j].x=x1;L1[j].y1=y1;L1[j].y2=y2;
           L2[j].y=y1;L2[j].x1=x1;L2[j].x2=x2;
           L1[j].flag=L2[j].flag=1;
           rcx[j]=x1;rcy[j]=y1;         j++;

           rcx[j]=x2;rcy[j]=y2;
           L1[j].x=x2;L1[j].y1=y1;L1[j].y2=y2;
           L2[j].y=y2;L2[j].x1=x1;L2[j].x2=x2;
           L1[j].flag=L2[j].flag=-1;     j++;
       }
      sort(L1,L1+j);//一下的代码冗长呀、呵呵
      sort(rcy,rcy+j);
      for(k=0,i=1;i<j;i++)
        if(rcy[i]!=rcy[k])
         rcy[++k]=rcy[i];
      build(0,k,1);
      len=0;
      for(i=0;i<j;i++)
        {
            pl=st[1].len;
            flag=L1[i].flag;
            update1(L1[i].y1,L1[i].y2,0,k,1);
            nl=st[1].len;
            len+=pl>nl?pl-nl:nl-pl;
        }
      sort(L2,L2+j);
      sort(rcx,rcx+j);
      for(k=0,i=1;i<j;i++)
        if(rcx[i]!=rcx[k])
         rcx[++k]=rcx[i];
      build(0,k,1);
      //len=0;
      for(i=0;i<j;i++)
        {
            pl=st[1].len;
            flag=L2[i].flag;
            update2(L2[i].x1,L2[i].x2,0,k,1);
            nl=st[1].len;
            len+=pl>nl?pl-nl:nl-pl;
        }
        printf("%d\n",len);
    }
    return 0;
}