HDU1255 覆盖的面积 扫描线

好久没写扫描线了，今天想拿些数据结构题目练手，在题目分类里面看到了这题，就拍上了。

扫描线排序离散化，线段树的区间代表y的脚标，然后对y进行离散化，二分查找对应的y。

val用来记录区间被覆盖的次数，严格O(nlogn)的访问，然后直接扫描一遍就可以了。

对精度要求不高，原来数组开小了，eps设成1e-8就WA了，不知道是数据开到1000不够还是eps太小。

 

我的代码：

 

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;const double eps=1e-6;const int MAX=10100;int n,len;double sum;struct Node{int l,r,val;double area; };Node seg[4*MAX];struct Line{//each sub contains two lines, left and right.double x,y1,y2;bool s;//true- left, false - rightbool operator<(const Line& line) const{return x<line.x;}};Line line[2*MAX]; double line_y[2*MAX];int Read(double num){int left,mid,right;left=0;right=len-1;while(left<=right){mid=(left+right)/2;if(fabs(num-line_y[mid])<eps)return mid;if(num-line_y[mid]>eps)left=mid+1;elseright=mid-1; }return -1;} void Init(int k,int l,int r)//clear{seg[k].l=l;seg[k].r=r;seg[k].val=0;seg[k].area=0;if(l+1<r){int mid=(l+r)/2;Init(2*k,l,mid);Init(2*k+1,mid,r); } }void Update(int k){if(seg[k].val>1)seg[k].area=line_y[seg[k].r]-line_y[seg[k].l];//calculate area else if(seg[k].r-seg[k].l==1)seg[k].area=0;elseseg[k].area=seg[2*k].area+seg[2*k+1].area; }void Insert(int k,int l,int r){if(seg[k].r-seg[k].l==1){seg[k].val++;Update(k);return ; }int mid=(seg[k].l+seg[k].r)/2;if(l<mid)Insert(k*2,l,r);if(r>mid)Insert(k*2+1,l,r);Update(k); }void Remove(int k,int l,int r){if(seg[k].r-seg[k].l==1){seg[k].val--;Update(k);return ; }int mid=(seg[k].l+seg[k].r)/2;if(l<mid)Remove(2*k,l,r);if(r>mid)Remove(2*k+1,l,r);Update(k); }void Add(double x1,double y1,double x2,double y2,int i)//add (x1,y1)-(x2,y2) line set.{int k=2*i;line[k].x=x1;line[k].y1=y1;line[k].y2=y2;line[k].s=1;line[k+1].x=x2;line[k+1].y1=y1;line[k+1].y2=y2;line[k+1].s=0;line_y[k]=y1;line_y[k+1]=y2; }void run(){//when input the cubes, use add() to add them.//then sort the array line and line_y.//in here, n is double of the number of cubes.//left and right.//after that, we delete the epactal lines in the set.//then insert the lines into segment tree.//let len is the new range of lines.n*=2; sort(line,line+n);sort(line_y,line_y+n);len=1;for(int i=1;i<n;i++)if(line_y[i]!=line_y[i-1])line_y[len++]=line_y[i];Init(1,0,len-1);sum=0.0;int l,r;//s=delta_x*delta_yfor(int i=0;i<n-1;i++){l=Read(line[i].y1);r=Read(line[i].y2);if(line[i].s)Insert(1,l,r);elseRemove(1,l,r);sum+=seg[1].area*(line[i+1].x-line[i].x); }printf("%.2f/n",sum);}int main(){double x1,x2,y1,y2;int t;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=0;i<n;i++){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);Add(x1,y1,x2,y2,i);}run();}return 0;}