扫描线，矩形面积求并hdu 1542

最新推荐文章于 2019-03-13 16:50:40 发布

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该程序需要处理多个输入的地图数据，每个数据包含四个坐标，表示地图覆盖的矩形区域。目标是计算所有矩形覆盖的总面积，并精确到小数点后两位。输出包括测试用例编号和总探索面积。

Atlantis

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

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<set>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<cmath>
using namespace std;
const int INF=0x7f7f7f7f;
const int maxn=2e6+5;
const int MOD=10007;
struct node
{
    double l,r,h;//线段的左右端点 以及高度
    int flag;
}a[205];
double x1,y1,x2,y2;
vector<double> vx;
bool cmp(node a,node b)
{
    return a.h<b.h;
}
struct tree
{
    int cnt;
    double sum;
}b[205<<2];
void up(int l,int r,int rt)
{
    if(b[rt].cnt)
    {
        b[rt].sum=vx[r]-vx[l-1];
        //cout<<r+1<<' '<<l<<endl;
    }
    else if(l==r)
    {
        b[rt].sum=0;
    }
    else b[rt].sum=b[rt<<1].sum+b[rt<<1|1].sum;

}
void update(int l,int r,int rt,int ll,int rr,int c)
{
    if(ll<=l&&r<=rr)
    {
        b[rt].cnt+=c;
        up(l,r,rt);
        return ;
    }
    int mid=(l+r)>>1;
    if(ll<=mid) update(l,mid,rt<<1,ll,rr,c);
    if(rr>mid) update(mid+1,r,rt<<1|1,ll,rr,c);
    up(l,r,rt);
}
int n;
int main()
{
    int cas=1;
    double ans;
    while(scanf("%d",&n)!=EOF&&n)
    {
        vx.clear();
        ans=0;
        for(int i=1;i<=n;++i)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            a[i].l=x1;
            a[i].r=x2;
            a[i].h=y1;
            a[i].flag=1;

            a[i+n].l=x1;
            a[i+n].r=x2;
            a[i+n].h=y2;
            a[i+n].flag=-1;
            vx.push_back(x1);
            vx.push_back(x2);
        }
        printf("Test case #%d\n",cas++);
        sort(a+1,a+1+2*n,cmp);
        sort(vx.begin(),vx.end());
        for(int i=1;i<vx.size();i++)
            if(vx[i]==vx[i-1]) vx.erase(i+vx.begin());//去重
        memset(b,0,sizeof b);
        for(int i=1;i<=n*2;++i)
        {
            int l=lower_bound(vx.begin(),vx.end(),a[i].l)-vx.begin()+1;
            int r=lower_bound(vx.begin(),vx.end(),a[i].r)-vx.begin();//区间左开右闭;
            update(1,vx.size(),1,l,r,a[i].flag);
            ans+=b[1].sum*(a[i+1].h-a[i].h);
        }
        printf("Total explored area: %.2f\n\n",ans);
    }
}