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;}