POJ 1177 Picture

最新推荐文章于 2025-09-07 00:18:35 发布

weixin_30501857

最新推荐文章于 2025-09-07 00:18:35 发布

阅读量85

点赞数

CC 4.0 BY-SA版权

文章标签： php

原文链接：http://www.cnblogs.com/372465774y/archive/2012/07/24/2606497.html

本文介绍了一种通过扫描线算法来计算多个矩形粘贴在墙上后形成的总面积边界长度（即周长）的方法。该算法利用了线段树的数据结构进行区间覆盖标记，并通过更新线段树中的覆盖状态来计算最终的周长。

Picture

Time Limit: 2000MS		Memory Limit: 10000K
Total Submissions: 8215		Accepted: 4347

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

Source

IOI 1998

http://acm.hdu.edu.cn/showproblem.php?pid=1828

//两个地方都试了，数据都不完整噢

//比如我开始交的代码是错的、还是A了

//继续扫描线、成段更新呀、是门学问

//不过这次代码感觉长度可以缩减好些的、好多语句就是一样的

//在函数里调用指针、这样代码长度可以减少好多

#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;
}