hdu 1542 扫描线模板题

https://vjudge.net/problem/HDU-1542

直接离散化+扫描线即可

主要是复习一下代码细节

#include<algorithm>
#include<iostream>
#include<iomanip>
#include<string.h>
#include<cstdlib>
#include<sstream>
#include<cstdio>
#include<cmath>
#define endl '
'
#define ll long long
#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(ii,a,b) for(register int ii=a;ii<=b;++ii)
#define forn(ii,now) for(register int ii=head[now];ii;ii=e[ii].next)
using namespace std;
const int maxn=1e3+10,maxm=2e6+10;
const ll INF=0x3f3f3f3f3f3f3f3f;
int casn,n,m,k;
double dis[maxn];
struct node{
  double x,y1,y2;int tag;
  bool operator <(node &other) const{return x<other.x;}
}seg[maxn];
class segtree{public:
#define nd  node[now]
#define ndl node[now<<1]
#define ndr node[now<<1|1]
    struct segnode {
        int l,r,tag;double dis;
        inline int mid(){return (r+l)>>1;}
        inline int len(){return r-l+1;}
    }node[maxn<<2|3];
    inline void update(int now){
      if(nd.tag) nd.dis=dis[nd.r+1]-dis[nd.l];
      else if(nd.len()==1) nd.dis=0;
      else nd.dis=ndl.dis+ndr.dis;
    }
    void maketree(int s,int t,int now=1){
        nd={s,t,0,0};
        if(s==t) return ;
        maketree(s,nd.mid(),now<<1);
        maketree(nd.mid()+1,t,now<<1|1);
    }
    void update(int s,int t,int x,int now=1){
      if(s<=nd.l&&t>=nd.r) {
        nd.tag+=x;update(now);
        return ;
      }
      if(s<=ndl.r) update(s,t,x,now<<1);
      if(t>ndl.r) update(s,t,x,now<<1|1);
      update(now);
    }
}tree;
int main() {IO;cout<<fixed<<setprecision(2);
  while((cin>>n)&&n){
    m=0;
    rep(i,1,n){
        double a,b,c,d;cin>>a>>b>>c>>d;
        dis[++m]=b;seg[m]={a,b,d,1};
        dis[++m]=d;seg[m]={c,b,d,-1};
    }
    sort(dis+1,dis+m+1);
    sort(seg+1,seg+m+1);
    int cnt=unique(dis+1,dis+1+m)-dis-1;
    tree.maketree(1,cnt);
    double ans=0;
    rep(i,1,m-1){
        int l=lower_bound(dis+1,dis+1+cnt,seg[i].y1)-dis;
        int r=lower_bound(dis+1,dis+1+cnt,seg[i].y2)-dis;
        r--;
        if(l<=r) tree.update(l,r,seg[i].tag);
        ans+=tree.node[1].dis*(seg[i+1].x-seg[i].x);
    }
    cout<<"Test case #"<<++casn<<endl;
    cout<<"Total explored area: "<<ans<<endl<<endl;
  }
}
原文地址:https://www.cnblogs.com/nervendnig/p/10873341.html