POJ 1151 Atlantis 线段树+离散化+扫描线

这次是求矩形面积并  

/*
Problem: 1151        User: 96655
Memory: 716K        Time: 0MS
Language: G++        Result: Accepted
*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
using namespace std;
const int maxn=110;
double y[maxn*2];
struct Line
{
    int co;
    double x,y1,y2;
    void fun(double a,double b,double c,int d)
    {
        x=a;
        y1=b;
        y2=c;
        co=d;
    }
    bool operator<(const Line &e)const
    {
        return x<e.x;
    }
}line[maxn*2];
struct Node
{
    double s,t,len;
    int co;
    void change(int o)
    {
        co+=o;
        if(co==0) len=0;
        else len=t-s;
    }
};
struct Segtree
{
    Node tree[maxn*8];
    void build(int l,int r,int o)
    {
       tree[o].s=y[l];tree[o].t=y[r];
       tree[o].co=0;tree[o].len=0;
       if(l+1<r)
       {
          int m=(l+r)>>1;
          build(l,m,o*2);
          build(m,r,o*2+1);
       }
    }
    void pushup(int o)
    {
      tree[o].len=tree[o*2].len+tree[o*2+1].len;
    }
    void update(int l,int r,int o,Line e)
    {
        if(l+1==r)
        {
           tree[o].change(e.co);
           return;
        }
        int m=(l+r)>>1;
        if(e.y1<tree[o*2].t)update(l,m,o*2,e);
        if(e.y2>tree[o*2+1].s)update(m,r,o*2+1,e);
        pushup(o);
    }
}seg;
int main()
{
    int n,ncase=0;
    while(~scanf("%d",&n),n)
    {
        int cnt=0;
        for(int i=0;i<n;i++)
        {
            double x1,x2,y1,y2;
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            line[cnt].fun(x1,y1,y2,1);
            y[cnt++]=y1;
            line[cnt].fun(x2,y1,y2,-1);
            y[cnt++]=y2;
        }
        sort(line,line+cnt);
        sort(y,y+cnt);
        int d=1;
        for(int i=1;i<cnt;i++)
           if(y[i]!=y[i-1])y[d++]=y[i];
        double ans=0;
        seg.build(0,d-1,1);
        seg.update(0,d-1,1,line[0]);
        for(int i=1;i<cnt;i++)
        {
           ans+=(line[i].x-line[i-1].x)*seg.tree[1].len;
           seg.update(0,d-1,1,line[i]);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n",++ncase,ans);
    }
    return 0;
}