K - Atlantis (线段树+扫描线）

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.

InputThe 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.OutputFor 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

#define lson i<<1,l,m
#define rson i<<1|1,m+1,r
const int maxn=222;
double x[maxn];
struct node
{
    double l,r,h;//左右坐标，高度
    int d;//标记上位边还是下位边
    node() {}
    node(double l,double r,double h,int d):l(l),r(r),h(h),d(d) {}
    bool operator < (const node &a)const
    {
        return h<a.h;
    }
} line[maxn];
int cnt[maxn<<2];
double sum[maxn<<2];
void pushup(int i,int l,int r)
{
    if(cnt[i])
        sum[i]=x[r+1]-x[l];
    else
        sum[i]=sum[i<<1]+sum[i<<1|1];
}
//       update(l,r,line[i].d,1,1,k-1);
void update(int ql,int qr,int v,int i,int l,int r)
{
    if(ql<=l && qr>=r)
    {
        cnt[i]+=v;
        pushup(i,l,r);
        return ;
    }
    int m=(l+r)>>1;
    if(ql<=m) update(ql,qr,v,lson);
    if(qr>m)  update(ql,qr,v,rson);
    pushup(i,l,r);
}
int main()
{
    int q;
    int kase=0;
    while(cin>>q&&q)
    {
        memset(cnt,0,sizeof(cnt));    //
        memset(sum,0,sizeof(sum));    //存储线段
        int n=0,m=0;
        for(int i=1; i<=q; i++)
        {
            double x1,y1,x2,y2;
            scanf("%lf %lf %lf %lf",&x1,&y1,&x2,&y2);
            x[++n]=x1;
            x[++n]=x2;
            line[++m] = node(x1,x2,y1,1);
            line[++m] = node(x1,x2,y2,-1);
        }
        sort(x+1,x+1+n);
        sort(line+1,line+1+m);
        int k=1;
        k=unique(x+1,x+n+1)-x-1;
        double ans=0.0;
        for(int i=1; i<m; i++)
        {

            int l=lower_bound(x+1,x+k+1,line[i].l)-x;
            int r=lower_bound(x+1,x+k+1,line[i].r)-x;
            r--;
            if(l<=r) update(l,r,line[i].d,1,1,k-1);
            ans+=sum[1]*(line[i+1].h-line[i].h);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n",++kase,ans);
    }
}

原文地址：https://www.cnblogs.com/Shallow-dream/p/11756311.html