Atlantis

Atlantis

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18287    Accepted Submission(s): 7417

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

2 10 10 20 20 15 15 25 25.5 0

 

Sample Output

Test case #1 Total explored area: 180.00

 

Source

Mid-Central European Regional Contest 2000

 

Recommend

linle   |   We have carefully selected several similar problems for you:  1828 1255 1166 1540 1754 

 

 离散化+线段树+扫描线

#include <bits/stdc++.h>
#define maxn 10005
struct Line
{
    double x,y_down,y_up;
    int flag;
    bool operator <(const Line &a)
    {
        return x<a.x;
    }
}line[maxn];
struct Tree
{
    double y_down,y_up;
    double x;
    int cover;//用以表示加进线段树的线段次数
    bool flag;
}tree[maxn*100];
int n;
double xx1,yy1,xx2,yy2;
int cnt=0;
double y[2*maxn];
void build(int l,int r,int rt)
{
    tree[rt].x=-1;
    tree[rt].cover=0;
    tree[rt].y_down=y[l];
    tree[rt].y_up=y[r];
    tree[rt].flag=false;
    if(l+1==r)
    {
        tree[rt].flag=true;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt*2);
    build(mid,r,rt*2+1);
}
double insert_tree(int rt,double x,double l,double r,int flag)//flag表示是左边还是右边
{
    if(r<=tree[rt].y_down||l>=tree[rt].y_up)
    {
        return 0;
    }
    if(tree[rt].flag)
    {
        if(tree[rt].cover>0)
        {
            double temp_x=tree[rt].x;
            double ans=(x-temp_x)*(tree[rt].y_up-tree[rt].y_down);
            tree[rt].x=x;
            tree[rt].cover+=flag;
            return ans;
        }
        else
        {
            tree[rt].cover+=flag;
            tree[rt].x=x;
            return 0;
        }
    }
    double ans1,ans2;
    ans1=insert_tree(rt*2,x,l,r,flag);
    ans2=insert_tree(rt*2+1,x,l,r,flag);
    return ans1+ans2;
}
int main()
{
    int n,i;
    int cas=0;
    while(std::cin>>n&&n)
    {
        cnt=1;
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&xx1,&yy1,&xx2,&yy2);
            y[cnt]=yy1;
            line[cnt].x=xx1;
            line[cnt].y_down=yy1;
            line[cnt].y_up=yy2;
            line[cnt].flag=1;
            cnt++;
            y[cnt]=yy2;
            line[cnt].x=xx2;
            line[cnt].y_down=yy1;
            line[cnt].y_up=yy2;
            line[cnt].flag=-1;
            cnt++;
        }
        std::sort(y+1,y+cnt);
        std::sort(line+1,line+cnt);
        build(1,cnt-1,1);
        double ans=0;
        for(int i=1;i<cnt;i++)
        {    //std::cout<<"gg"<<std::endl;
            ans+=insert_tree(1,line[i].x,line[i].y_down,line[i].y_up,line[i].flag);
        }
         printf("Test case #%d
Total explored area: %.2f

", ++cas, ans);
    }
    return 0;
}