hdu 1255 线段树 求矩形面积的交

在 hdu1542 的基础上 再加一个calen2函数

#include<stdio.h>
#include<stdlib.h>
#include <algorithm>
#define ls o<<1
#define rs o<<1|1
#define N 2010
void fre(){freopen("t.txt","r",stdin);}
using namespace std;
struct node{
    double x,y1,y2;
    int f;
}Line[N];
struct node1{
    double lf,rf,cnt,cnt2;
    int l,r,c;
}tree[N*3];
double y[N];
int cmp(node a,node b)
{
    return a.x < b.x;
}
int cmp1(double a,double b)
{
    return a < b;
}
void bulid(int o,int l,int r)
{
    int m = l+(r-l)/2;
    tree[o].c=0;tree[o].cnt=0;
    tree[o].cnt2=0;
    tree[o].l=l;
    tree[o].r=r;
    tree[o].lf=y[l];
    tree[o].rf=y[r];
    if(l+1==r) return;//注意返回条件
    bulid(ls,l,m);
    bulid(rs,m,r);//注意这里不用mid+1
}
void calen(int t)
{
    if(tree[t].c>0)
    {
        tree[t].cnt=tree[t].rf-tree[t].lf;
        return ;
    }
    if(tree[t].l+1==tree[t].r) tree[t].cnt=0;
    else tree[t].cnt=tree[2*t].cnt+tree[2*t+1].cnt;
}
void calen2(int t)//新加的函数
{
    if(tree[t].c > 1)
    {
        tree[t].cnt2 = tree[t].rf - tree[t].lf;
        return;
    }
    if(tree[t].l+1==tree[t].r) tree[t].cnt2=0;
    else if(tree[t].c == 1)
    {
        tree[t].cnt2 = tree[2*t].cnt + tree[2*t+1].cnt;
    }
    else tree[t].cnt2 = tree[2*t].cnt2 + tree[2*t+1].cnt2;
}
void updata(int t,node e)
{
    if(e.y1 <= tree[t].lf && e.y2>=tree[t].rf)
    {
        tree[t].c+=e.f;
        calen(t);
        calen2(t);
        return;
    }
    if(e.y1 < tree[2*t].rf ) updata(2*t,e);
    if(e.y2 > tree[2*t+1].lf) updata(2*t+1,e);
    calen(t);
    calen2(t);
}
int main()
{
    //fre();
    int i,n,kase,t;
    scanf("%d",&kase);
    double x1,y1,x2,y2,ans;
    while(kase--)
    {
        scanf("%d",&n);
        t=1;

        for(i=1;i<=n;i++)
        {
            scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
            Line[t].x=x1;
            Line[t].y1=y1;
            Line[t].y2=y2;
            Line[t].f=1;
            y[t]=y1;

            Line[t+1].x=x2;
            Line[t+1].y1=y1;
            Line[t+1].y2=y2;
            Line[t+1].f=-1;
            y[t+1]=y2;
            t+=2;
        }
        sort(Line+1,Line+t,cmp);
        sort(y+1,y+t,cmp1);
        bulid(1,1,t-1);

        ans=0;
        for(i=1;i<t;i++)
        {
            ans+=tree[1].cnt2*(Line[i].x - Line[i-1].x);
            updata(1,Line[i]);
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}