hdu 1255 覆盖的面积 矩形D层以上重叠面积

矩形面积并,求D层以上重叠的面积

离散化后扫描线 4000ms低效率水过

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const double EP=1e-8;
const int maxn=2005;
const int D=2;
int n, cas=1;
double hash[maxn];
struct nod{
    double x;
    int c, id;
}a[maxn];

struct node{
    int lef, rig, mid, cover;
}seg[4*maxn];

struct line{
    int x1, x2, flag;    //flag==1 down
    double y;
}l[maxn];

void make_tree(int num, int lef, int rig){
    seg[num].lef=lef;
    seg[num].rig=rig;
    seg[num].mid=(lef+rig)>>1;
    seg[num].cover=0;
    if(lef+1!=rig){
        make_tree(num<<1, lef, seg[num].mid);
        make_tree(num*2+1, seg[num].mid, rig);
    }
}

void insert(int num, int lef, int rig, int cover){
    if(seg[num].cover!=-1&&seg[num].lef==lef&&seg[num].rig==rig){
        seg[num].cover+=cover;
        return;
    }
    if(seg[num].cover>0){
        insert(num*2, seg[num*2].lef, seg[num*2].rig, seg[num].cover);
        insert(num*2+1, seg[num*2+1].lef, seg[num*2+1].rig, seg[num].cover);
        seg[num].cover=-1;
    }
    seg[num].cover=-1;
    if(rig<=seg[num].mid)
        insert(num*2, lef, rig, cover);
    else if(lef>=seg[num].mid)
        insert(num*2+1, lef, rig, cover);
    else {
        insert(num*2, lef, seg[num].mid, cover);
        insert(num*2+1, seg[num].mid, rig, cover);
    }
}

double cal(int num, int cover){
    if(seg[num].cover>=cover)
    return hash[seg[num].rig]-hash[seg[num].lef];
    if(seg[num].lef+1==seg[num].rig){
        return 0;
    }
    return cal(num*2, cover)+cal(num*2+1, cover);
}

bool cmpx(nod p1, nod p2){
    return p1.x<p2.x;
}
bool cmpid(nod p1, nod p2){
    return p1.id<p2.id;
}
bool cmpy(line l1, line l2){
    return l1.y<l2.y;
}

void myhash(){
    int i;
    sort(a, a+2*n, cmpx);
    hash[0]=a[0].x;
    a[0].c=0;
    for(i=1; i<2*n; i++){
        if(a[i].x-a[i-1].x<EP)
        a[i].c=a[i-1].c;
        else {
            a[i].c=a[i-1].c+1;
            hash[a[i].c]=a[i].x;
        }
    }
    sort(a, a+2*n, cmpid);
    for(i=0; i<n; i++){
        l[i*2].x1=l[i*2+1].x1=a[i*2].c;
        l[i*2].x2=l[i*2+1].x2=a[i*2+1].c;
    }
}

void init(){
    for(int i=0; i<n; i++){
        scanf("%lf%lf%lf%lf", &a[i*2].x, &l[i*2].y, &a[i*2+1].x, &l[i*2+1].y);
        l[i*2].flag=1;l[i*2+1].flag=-1;
        a[i*2].id=i*2;a[i*2+1].id=i*2+1;
    }
}

void solve(){
    make_tree(1, 0, 2*n);
    myhash();
    sort(l, l+2*n, cmpy);
    double ans=0;
    for(int i=0; i<2*n-1; i++){
        insert(1, l[i].x1, l[i].x2, l[i].flag);
        ans+=cal(1, D)*(l[i+1].y-l[i].y);
    }
    printf("%.2f\n", ans);
}

int main(){
    //freopen("1.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        init();
        solve();
    }
    return 0;
}


 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值