HDU1255 覆盖的面积(线段树+扫描线)

对于覆盖的一次的,我们维护一格变量,对于覆盖两次的,我们维护两个变量,分别是s1,s2,覆盖一次以上和覆盖两次以上。

通过这样的操作就可以直接根据儿子信息更新而不用维护一格pushdown操作

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=3e5+10;
const int mod=1e9+7;
struct node{
    int l,r,lazy;
    double s1,s2;
}tr[N];
vector<double> num;
struct qi{
    double x;
    double y,y2;
    int k;
    bool operator <(const qi& t) const{
        return x<t.x;
    }
}sg[N];
int find(double x){
    return lower_bound(num.begin(),num.end(),x)-num.begin()+1;
}
void build(int u,int l,int r){
    if(l==r){
        tr[u]=node{l,l,0,0,0};
    }
    else{
        tr[u]=node{l,r,0,0,0};
        int mid=l+r>>1;
        build(u<<1,l,mid);
        build(u<<1|1,mid+1,r);
    }
}
void pushup(int u){
    if(tr[u].lazy>=2){
        tr[u].s2=num[tr[u].r]-num[tr[u].l-1];
        tr[u].s1 =num[tr[u].r]-num[tr[u].l-1];
    }
    else if(tr[u].lazy==1){
        tr[u].s2=tr[u<<1].s1+tr[u<<1|1].s1;
        tr[u].s1 =num[tr[u].r]-num[tr[u].l-1];
    }
    else{
        if(tr[u].l==tr[u].r)
            tr[u].s2 = tr[u].s1 = 0;
        else{
            tr[u].s2= tr[u<<1].s2+tr[u<<1|1].s2;
            tr[u].s1 = tr[u<<1].s1+tr[u<<1|1].s1;
        }
    }
}
void modify(int u,int l,int r,int x){
    if(tr[u].l>=l&&tr[u].r<=r){
        tr[u].lazy+=x;
        pushup(u);
        return ;
    }
    int mid=tr[u].l+tr[u].r>>1;
    if(l<=mid)
        modify(u<<1,l,r,x);
    if(r>mid)
        modify(u<<1|1,l,r,x);
    pushup(u);
}
int main(){
    int n;
    int t;
    cin>>t;
    while(t--){
        cin>>n;
        num.clear();
        int i;
        double x1,x2,y,y2;
        int cnt=1;
        for(i=1;i<=n;i++){
               scanf("%lf%lf%lf%lf",&x1,&y,&x2,&y2);
               sg[cnt++]={x1,y,y2,1};
               sg[cnt++]={x2,y,y2,-1};
            num.push_back(y);
            num.push_back(y2);
        }
        sort(sg+1,sg+1+2*n);
        sort(num.begin(),num.end());
        num.erase(unique(num.begin(),num.end()),num.end());
        build(1,1,num.size()-1);
        double sum=0;
        for(i=1;i<=2*n;i++){
            if(i>1){
                sum+=tr[1].s2*(sg[i].x-sg[i-1].x);
            }
            modify(1,find(sg[i].y),find(sg[i].y2)-1,sg[i].k);
        }
        printf("%.2f
",sum);
    }
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/ctyakwf/p/12871872.html