hdu 1542 Atlantis(线段树 线性扫描）

Problem 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 file 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.

 

Sample Input

210 10 20 2015 15 25 25.50

 

Sample Output

Test case #1Total explored area: 180.00 

 

Source

Mid-Central European Regional Contest 2000

 

题意：给你n个矩形，求它们的面积，重复的不重复计算

思路：线段树的线性扫描，由于坐标可能很大，需要先离散化处理；

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int N=105;struct node1{    double x1,x2,h;    int f;}s[2*N];struct node2{    int l,r,cnt;    double len;}str[6*N];double a[2*N];bool cmp(node1 aa,node1 bb){    return aa.h<bb.h;}int find(double x,int l,int r){    while(l<=r)    {        int temp=(l+r)/2;        if(x==a[temp]) return temp;        else if(x<a[temp]) r=temp-1;        else l=temp+1;    }}void build(int l,int r,int n){    str[n].l=l;    str[n].r=r;    str[n].len=0;    str[n].cnt=0;    if(l==r)        return;    int temp=(l+r)/2;    build(l,temp,2*n);    build(temp+1,r,2*n+1);}void add(int n){    if(str[n].cnt)        str[n].len=a[str[n].r+1]-a[str[n].l];    else if(str[n].l==str[n].r)        str[n].len=0;    else        str[n].len=str[2*n].len+str[2*n+1].len;}void updata(int l,int r,int flag,int n){    if(l==str[n].l&&r==str[n].r)    {        str[n].cnt+=flag;        add(n);        return;    }    int temp=(str[n].l+str[n].r)/2;    if(r<=temp)        updata(l,r,flag,2*n);    else if(l>temp)        updata(l,r,flag,2*n+1);    else    {        updata(l,temp,flag,2*n);        updata(temp+1,r,flag,2*n+1);    }    add(n);}int main(){    int n;    int t=1;    while(~scanf("%d",&n))    {        if(n==0) break;        for(int i=0;i<n;i++)        {            double x1,y1,x2,y2;            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);            s[2*i].x1=x1;s[2*i].x2=x2;s[2*i].h=y1;s[2*i].f=1;            s[2*i+1].x1=x1;s[2*i+1].x2=x2;s[2*i+1].h=y2;s[2*i+1].f=-1;            a[2*i]=x1;a[2*i+1]=x2;        }        sort(s,s+2*n,cmp);        sort(a,a+2*n);        int len=1;        double temp=a[0];        for(int i=1;i<2*n;i++)            if(temp!=a[i])            {                a[len++]=a[i];                temp=a[i];            }        build(0,len-1,1);        double ans=0;        for(int i=0;i<2*n;i++)        {            int l=find(s[i].x1,0,len-1);            int r=find(s[i].x2,0,len-1)-1;            updata(l,r,s[i].f,1);            ans+=str[1].len*(s[i+1].h-s[i].h);        }        printf("Test case #%d\n",t++);        printf("Total explored area: %.2lf\n\n",ans);    }}