TOJ2712: Atlantis

小数据求面积并

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 Europe 2000

这个直接用了二分查询，但是复杂度还是蛮高的

#include<bits/stdc++.h>
using namespace std;
const int N=;
const double eps=1e-;
int n,vis[N],t;
double has[N];
struct line
{
    double l,r,y;
    int f;
    line(){}
    line(double l,double r,double y,int f):l(l),r(r),y(y),f(f){}
}L[N];
bool cmp(line a,line b)
{
    return a.y<b.y;
}
int get(double x)
{
    return lower_bound(has,has+t,x)-has;
}
int main()
{
    int ca=,i,j;
    double x1,y1,x2,y2;
    while(scanf("%d",&n),n)
    {
        t=;
        memset(vis,,sizeof vis);
        for(i=; i<n; i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            L[t]=line(x1,x2,y1,),has[t++]=x1;
            L[t]=line(x1,x2,y2,-),has[t++]=x2;
        }
        n*=;
        sort(L,L+n,cmp);
        sort(has,has+t);
        t=;
        for(i=;i<n;i++)
            if(fabs(has[i]-has[i-])>eps)has[t++]=has[i];
        double res=;
        for(i=; i<n; i++)
        {
            int l=get(L[i].l),r=get(L[i].r);
            double len=;
            for(j=;j<t-;j++)
                if(vis[j]>)len+=has[j+]-has[j];
            if(i)res+=len*(L[i].y-L[i-].y);
             for(j=l;j<r;j++)vis[j]+=L[i].f;
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n",++ca,res);
    }
    return ;
}