nbut线段树专题 J - Atlantis

Description

There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.

Input

The input consists of several test cases. Each test case starts with a line containing a single integer n (1 <= n <= 100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0 <= x1 < x2 <= 100000;0 <= y1 < y2 <= 100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area. 
The input file is terminated by a line containing a single 0. Don't process it.

Output

For each test case, your program should output one section. The first line of each section must be "Test case #k", where k is the number of the test case (starting with 1). The second one must be "Total explored area: a", where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point. 
Output a blank line after each test case.

借这题学会了扫描线的思想，这题题意是求矩形面积的并，对横轴建树，注意浮点数离散化，由于要用到离散前的坐标，可以用map来保存，而且map自带对key值排序功能，实现起来很方便。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
#include<iostream>
using namespace std;
const int maxn=210;//离散完了后内存消耗小了很多
struct node
{
 int l,r,add;
 double len;
}tree[maxn<<2];

struct L
{
 int flag;
 double l,r,h;
}line[maxn<<1];

double pos[maxn<<1];

int cmp(L a,L b)
{
 return a.h < b.h;
}

void build(int p,int l,int r)
{
 tree[p].l=l;
 tree[p].r=r;
 tree[p].add=0;
 tree[p].len=0.0;
 if(l==r)
 return ;
 int mid=(l+r)>>1;
 build(p<<1,l,mid);
 build(p<<1|1,mid+1,r);
}

void pushup(int p)
{
 if(tree[p].add)
 tree[p].len=(pos[tree[p].r+1]-pos[tree[p].l]);
 else if(tree[p].l == tree[p].r)
 tree[p].len=0;
 else
 tree[p].len=(tree[p<<1].len + tree[p<<1|1].len);
}

void update(int p,int l,int r,int val)
{
 if(l<=tree[p].l && r>=tree[p].r)
 {
 tree[p].add+=val;
 pushup(p);
 return ;
 }
 int mid=(tree[p].l + tree[p].r)>>1;
 if(r<=mid)
 update(p<<1,l,r,val);
 else if(l>mid)
 update(p<<1|1,l,r,val);
 else
 {
 update(p<<1,l,mid,val);
 update(p<<1|1,mid+1,r,val);
 }
 pushup(p);
}

int main()
{
 int n,p=1;
 while(~scanf("%d",&n) && n)
 {
 int i,j;
 int k=0,flag=0,cnt=0;
 double ans=0;
 double x1,y1,x2,y2;
 map<double,int>point;
 map<double,int>::iterator it;
 for(i=0;i<n;i++)
 {
 scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
 line[cnt].l=x1;
 line[cnt].r=x2;
 line[cnt].h=y1;
 line[cnt++].flag=1;//下边
 line[cnt].l=x1;
 line[cnt].r=x2;
 line[cnt].h=y2;
 line[cnt++].flag=-1;//上边
 if(point[x1]==0)
 point[x1]=++flag;
 if(point[x2]==0)
 point[x2]==++flag;//注意，这里的flag不是离散化后的对应坐标
 }
 build(1,1,flag);
 for(it=point.begin();it!=point.end();it++)
 {
 pos[++k]=it->first;//pos数组保存离散前的坐标
 it->second=k;//k才是离散前的坐标所对应的离散坐标
 }
 sort(line,line+cnt,cmp);
 for(i=0;i<cnt-1;i++)
 {
 int l=point[line[i].l];
 int r=point[line[i].r]-1;
 update(1,l,r,line[i].flag);
 ans+=(line[i+1].h-line[i].h)*tree[1].len;
 }
 printf("Test case #%d\n",p++);
 printf("Total explored area: %.2f \n\n",ans);
 }
 return 0;
}