HDU1542 Atlantis【线段树+扫描线+离散化】
Atlantis
http://acm.hdu.edu.cn/showproblem.php?pid=1542
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21633 Accepted Submission(s): 8576
Problem Description
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.
Input
The 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
The input file is terminated by a line containing a single 0. Don’t process it.
Output
For 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
2
10 10 20 20
15 15 25 25.5
0Sample Output
Test case #1
Total explored area: 180.00 Source
Mid-Central European Regional Contest 2000
题意
二维平面有n个平行于坐标轴的矩形,现在要求出这些矩形的总面积，重叠部分只能算一次。
思路
线段树+扫描线，cnt[i]表示i控制区间的下位边-上位边的值，如果值为-1表示它左、右孩子的cnt[]值不一致。sum[i]表示i控制的区间中cnt[]值大于0的区间的长度。注意，线段树中每个叶节点(控制区间[L,L])不是指x[L]坐标,而是指区间[x[L],x[L+1]].线段树中其他节点控制的区间[L,R],也是指的x坐标轴的第L个区间到第R个区间的范围,也就是x[L]到x[R+1]坐标的范围。
C++代码
#include
#include
using namespace std;
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
const int N=250;
struct Node{
double l,r,h;
int d;
Node(){}
Node(double l,double r,double h,double d):l(l),r(r),h(h),d(d){}
bool operator<(const Node &t)const
{
return h>1;
build(ls);
build(rs);
pushup(rt);
}
void pushdown(int l,int r,int rt)
{
if(cnt[rt]!=-1)
{
int m=(l+r)>>1;
cnt[rt<<1]=cnt[rt<<1|1]=cnt[rt];
sum[rt<<1]=(cnt[rt]?x[m+1]-x[l]:0);
sum[rt<<1|1]=(cnt[rt]?x[r+1]-x[m+1]:0);
}
}
void update(int L,int R,int C,int l,int r,int rt)
{
if(L<=l&&r<=R)
{
if(cnt[rt]!=-1)
{
cnt[rt]+=C;
sum[rt]=(cnt[rt]?x[r+1]-x[l]:0);
return;
}
}
pushdown(l,r,rt);
int m=(l+r)>>1;
if(L<=m)
update(L,R,C,ls);
if(R>m)
update(L,R,C,rs);
pushup(rt);
}
int main()
{
int q,t;
while(~scanf("%d",&q)&&q)
{
int n=0,m=0;
for(int i=1;i<=q;i++)
{
double x1,y1,x2,y2;
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
x[++n]=x1;//左端点
x[++n]=x2;//右端点
nodes[++m]=Node(x1,x2,y1,1);//下位边
nodes[++m]=Node(x1,x2,y2,-1);//上位边
}
sort(x+1,x+n+1);//对x坐标排序
sort(nodes+1,nodes+m+1); //对上位边、下位边进行排序
n=unique(x+1,x+n+1)-(x+1);//去除相同的x坐标，进行离散化
build(1,n-1,1);
double ans=0;
for(int i=1;i