HDOJ 题目1828 Picture （线段树+扫描线）

Picture

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3646    Accepted Submission(s): 1884

Problem 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.

Please process to the end of file.

 

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 2515 -4 24 140 -6 16 42 15 10 2230 10 36 2034 0 40 16

 

Sample Output

228

 

Source

IOI 1998

 

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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("%.lf\n",ans);      }  }