POJ 1151：Atlantis 线段树+扫描线
Atlantis
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 19374 | Accepted: 7358 |
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 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.
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.
Output a blank line after each test case.
Sample Input
2 10 10 20 20 15 15 25 25.5 0
Sample Output
Test case #1 Total explored area: 180.00
题意是给出诸多矩形的左下角坐标和右上角坐标，问这些矩形一共覆盖的面积。
看了真的是很久很久，终于把这道题的代码搞懂了，趁着对这道题有很多感觉，赶紧记下来。也通过这些题，发现线段树的应用范围真的是广。
实际上就是对每一段的x，求其y的长度，相乘就是对应的面积。最后把所有的面积加起来就是总面积。为了防止可能会发生重复加的面积，使用了线段树的结构。
首先是离散化，把y坐标上的浮点数 变成1 2 3 ---，然后使用对每一条边使用二分查找。
然后线段树中的每一个节点就代表了相应段的y的长度，和之前的线段树一样，线段树的思想就是我要使用哪一段的长度，我再去计算它更新它，这里面的len就代表了这个含义。
然后就是cover，cover就是只有为0的时候去更新它，因为不是0的时候代表着它依然被某个矩形覆盖着，所以不能更新成子节点的长度之和。
dele就是将原来的长度划出掉，cover--，判断为0就更新该点的len。
写到哪里算哪里，看着这段代码，觉得真的很奇妙。
代码：
#pragma warning(disable:4996)
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
double y[210];
struct li
{
double x,y1,y2;
bool bleft;
}lines[210];
struct no
{
int L,R;
double len;
int cover;
}tree[10000];
bool operator < (const li &n1,const li &n2)
{
return n1.x<n2.x;
}
template <class F,class T>
F bin_search(F s,F e,T val)
{
F L = s;
F R = e-1;
while(L<=R)
{
F mid = L + (R-L)/2;
if(!(*mid<val || val < *mid))
{
return mid;
}
else if(val < *mid)
{
R = mid -1;
}
else
{
L= mid + 1;
}
}
}
void buildtree(int root,int L,int R)
{
tree[root].L=L;
tree[root].R=R;
tree[root].len=0;
tree[root].cover=0;
if(L!=R)
{
int mid = (L+R)/2;
buildtree(root*2+1,L,mid);
buildtree(root*2+2,mid+1,R);
}
}
void inse(int root,int L,int R)
{
if(tree[root].L==L&&tree[root].R==R)
{
tree[root].cover++;
tree[root].len= y[R+1] - y[L];
return;
}
int mid = (tree[root].L + tree[root].R)/2;
if(R<=mid)
{
inse(root*2+1,L,R);
}
else if(L>=mid+1)
{
inse(root*2+2,L,R);
}
else
{
inse(root*2+1,L,mid);
inse(root*2+2,mid+1,R);
}
if(tree[root].cover==0)
{
tree[root].len = tree[root*2+1].len + tree[root*2+2].len;
}
}
void dele(int root,int L,int R)
{
if(tree[root].L == L && tree[root].R == R)
{
tree[root].cover--;
if(tree[root].cover==0)
{
if(tree[root].L == tree[root].R)
{
tree[root].len=0;
}
else
{
tree[root].len = tree[root*2+1].len + tree[root*2+2].len;
}
}
return;
}
int mid = (tree[root].L + tree[root].R)/2;
if(R<=mid)
{
dele(root*2+1,L,R);
}
else if(L>=mid+1)
{
dele(root*2+2,L,R);
}
else
{
dele(root*2+1,L,mid);
dele(root*2+2,mid+1,R);
}
if(tree[root].cover==0)
{
tree[root].len = tree[root*2+1].len + tree[root*2+2].len;
}
}
int n;
int main()
{
int i,yc,lc,t;
double x1,x2,y1,y2;
t=0;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
t++;
yc=0;
lc=0;
for(i=1;i<=n;i++)
{
scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
y[yc++]=y1;
y[yc++]=y2;
lines[lc].x=x1;
lines[lc].y1=y1;
lines[lc].y2=y2;
lines[lc].bleft=true;
lc++;
lines[lc].x=x2;
lines[lc].y1=y1;
lines[lc].y2=y2;
lines[lc].bleft=false;
lc++;
}
sort(lines,lines+lc);
sort(y,y+yc);
yc=unique(y,y+yc)-y;
buildtree(0,0,yc-1-1);
double Area=0;
for(i=0;i<lc-1;i++)
{
int L=bin_search(y,y+yc,lines[i].y1)-y;
int R=bin_search(y,y+yc,lines[i].y2)-y;
if(lines[i].bleft)
{
inse(0,L,R-1);
}
else
{
dele(0,L,R-1);
}
Area += tree[0].len*(lines[i+1].x-lines[i].x);
}
printf("Test case #%d\n",t);
printf("Total explored area: %.2lf\n",Area);
printf("\n");
}
//system("pause");
return 0;
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