首页 > 代码库 > 线段树扫描线（hdu1542）
线段树扫描线（hdu1542）
There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of Atlantis. Your friend Bill has to know the total area for which maps exist. You (unwisely) volunteered to write a program that calculates this quantity.
InputThe input file consists of several test cases. Each test case starts with a line containing a single integer n (1<=n<=100) of available maps. The n following lines describe one map each. Each of these lines contains four numbers x1;y1;x2;y2 (0<=x1<x2<=100000;0<=y1<y2<=100000), not necessarily integers. The values (x1; y1) and (x2;y2) are the coordinates of the top-left resp. bottom-right corner of the mapped area.
The input file is terminated by a line containing a single 0. Don’t process it.OutputFor each test case, your program should output one section. The first line of each section must be “Test case #k”, where k is the number of the test case (starting with 1). The second one must be “Total explored area: a”, where a is the total explored area (i.e. the area of the union of all rectangles in this test case), printed exact to two digits to the right of the decimal point.
Output a blank line after each test case.
Sample Input
210 10 20 2015 15 25 25.50Sample Output
Test case #1Total explored area: 180.00
就是给定你2若干个矩形，求覆盖的面积。
看数据范围，就要用离散化来做，我们可以使用扫描线，沿着x轴扫过去，将每一个x坐标所对应的y的上下界用线段树表示出来，当一个矩形入边加入的时候就+1，否则就-1，然后算出面积就可以了。
具体看代码，我写了一些较为详细的注释。
#pragma GCC optimize("O2")#include<iostream>#include<cmath>#include<cstdlib>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<ctype.h>#define ls (rt<<1)#define rs (rt<<1|1)#define mid (l+(r-l)/2)#define maxn 500#define LL long longusing namespace std;struct TREE{int l,r,c;double cnt,lf,rf;}tr[maxn*4];//l,r存线段树的下标，c存+1-1，就是这颗扫描线扫过的矩形的数量,cnt存答案，也就是一段区间，lf,rf为区间的左右界 struct Line{double x,y1,y2;int c;}line[maxn];int m=0,n;double y[maxn];bool cmp(Line a,Line b){return a.x<b.x;}void build(int l,int r,int rt)//离散化，坐标一一对应 { tr[rt].l=l;tr[rt].r=r; tr[rt].cnt=tr[rt].c=0; tr[rt].lf=y[l],tr[rt].rf=y[r]; if(l+1==r) return;//左闭右开的区间 build(l,mid,ls); build(mid,r,rs);//同上所以不是mid+1 }//说一下区间为什么要 左闭右开，因为这样在计算面积的时候就不用+1-1，就不会出现细节上的错误 void getlen(int rt){ if(tr[rt].c) { tr[rt].cnt=tr[rt].rf-tr[rt].lf; // tr[rt].c=0;//注意这里不能清空，应该在下一条出边加入的时候才能清空 return ; } if(tr[rt].l+1==tr[rt].r) tr[rt].cnt=0;//如果是一个节点而不是区间 else tr[rt].cnt=tr[ls].cnt+tr[rs].cnt; return ;}void update(int rt,Line e){ if(e.y1==tr[rt].lf&&e.y2==tr[rt].rf) { tr[rt].c+=e.c; getlen(rt); return ; } if(e.y2<=tr[ls].rf) update(ls,e);//左儿子和右儿子要分开计算 else if(e.y1>=tr[rs].lf) update(rs,e); else { Line tmp=e; tmp.y2=tr[ls].rf; update(ls,tmp); tmp=e; tmp.y1=tr[rs].lf; update(rs,tmp); } getlen(rt);}void input(){ double x1,x2,y1,y2; for(int i=1;i<=n;i++) { cin>>x1>>y1>>x2>>y2; line[++m].x=x1;line[m].y1=y1;line[m].y2=y2;line[m].c=1; y[m]=y1; line[++m].x=x2;line[m].y1=y1;line[m].y2=y2;line[m].c=-1; y[m]=y2; } sort(line+1,line+m+1,cmp); sort(y+1,y+m+1);//坐标离散化 }void init()//记得初始化 { memset(tr,0,sizeof(tr)); m=0; memset(line,0,sizeof(line)); memset(y,0,sizeof(y));}int main(){ int kase=0; while(cin>>n&&n) { init(); input(); build(1,m,1); update(1,line[1]);//先加入第一条线 double ans=0; for(int i=2;i<=m;i++) { ans+=tr[1].cnt*(line[i].x-line[i-1].x);//标准的面积计算公式 update(1,line[i]); } printf("Test case #%d\n", ++kase); printf("Total explored area: %.2lf\n\n",ans); } return 0;}/*21 1 3 32 2 4 4*/
线段树扫描线（hdu1542）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。