HDU 1264 Counting Squares (线段树-扫描线-矩形面积并)
2024-09-24 04:46:26
版权声明:欢迎关注我的博客。本文为博主【炒饭君】原创文章,未经博主同意不得转载 https://blog.csdn.net/a1061747415/article/details/25471349
Problem A : Counting Squares
pid=1264" rel="nofollow">From:HDU, 1264
Problem Description
Your input is a series of rectangles, one per line. Each rectangle is specified as two points(X,Y) that specify the opposite corners of a rectangle. All coordinates will be integers in the range 0 to 100. For example, the line
5 8 7 10
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.
5 8 7 10
specifies the rectangle who's corners are(5,8),(7,8),(7,10),(5,10).
If drawn on graph paper, that rectangle would cover four squares. Your job is to count the number of unit(i.e.,1*1) squares that are covered by any one of the rectangles given as input. Any square covered by more than one rectangle should only be counted once.
Input
The input format is a series of lines, each containing 4 integers. Four -1's are used to separate problems, and four -2's are used to end the last problem. Otherwise, the numbers are the x-ycoordinates of two points that are opposite corners of
a rectangle.
a rectangle.
Output
Your output should be the number of squares covered by each set of rectangles. Each number should be printed on a separate line.
Sample Input
5 8 7 10
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2
6 9 7 8
6 8 8 11
-1 -1 -1 -1
0 0 100 100
50 75 12 90
39 42 57 73
-2 -2 -2 -2
Sample Output
8
10000
10000
Source
Recommend
JGShining
题目大意:
给定你一些矩形左下右上角坐标点。或者左上右下坐标点。求这些矩形的面积并。
解题思路:
利用线段树扫描线的知识。此题不须要离散化。
#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
struct node{
int x,y1,y2,c;
node(int x0=0,int y10=0,int y20=0,int c0=0){
x=x0;y1=y10;y2=y20;c=c0;
}
friend bool operator < (node a,node b){
if(a.x!=b.x) return a.x<b.x;
else if(a.y1!=b.y1) return a.y1<b.y1;
else if(a.y2!=b.y2) return a.y2<b.y2;
else return a.c>b.c;
}
};
const int maxh=110;
struct tree{
int l,r,c,lz;
}a[maxh*4];
vector <node> v;
bool input(){
int a,b,c,d;
v.clear();
while(scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF){
if(a==-1 && b==-1 && c==-1 && d==-1) return true;
if(a==-2 && b==-2 && c==-2 && d==-2) return false;
v.push_back(node( min(a,c), min(b,d) , max(b,d) ,1));
v.push_back(node( max(a,c), min(b,d) , max(b,d) ,-1));
}
}
void build(int l,int r,int k){
a[k].l=l;
a[k].r=r;
a[k].c=0;
a[k].lz=0;
if(l+1<r){
int mid=(l+r)/2;
build(l,mid,2*k);
build(mid,r,2*k+1);
}
}
void pushdown(int k){
if(a[k].lz!=0 && a[k].l+1<a[k].r ){
a[2*k].lz+=a[k].lz;
a[2*k+1].lz+=a[k].lz;
a[2*k].c+=a[k].lz;
a[2*k+1].c+=a[k].lz;
a[k].lz=0;
}
}
void insert(int l,int r,int k,int c){
if(l<=a[k].l && a[k].r<=r){
a[k].lz+=c;
a[k].c+=c;
}else{
pushdown(k);
int mid=(a[k].l+a[k].r)/2;
if(r<=mid) insert(l,r,2*k,c);
else if(l>=mid) insert(l,r,2*k+1,c);
else{
insert(l,mid,2*k,c);
insert(mid,r,2*k+1,c);
}
}
}
int query(int l,int r,int k){
pushdown(k);
if(l<=a[k].l && a[k].r<=r){
if(a[k].c>0) return r-l;
else{
if(a[k].l+1==a[k].r) return 0;
else {
int mid=(a[k].l+a[k].r)/2;
return query(l,mid,2*k) + query(mid,r,2*k+1) ;
}
}
}else{
int mid=(a[k].l+a[k].r)/2;
if(r<=mid) return query(l,r,2*k);
else if(l>=mid) return query(l,r,2*k+1);
else{
return query(l,mid,2*k) + query(mid,r,2*k+1) ;
}
}
}
void solve(){
build(0,maxh,1);
sort(v.begin(),v.end());
insert(v[0].y1,v[0].y2,1,v[0].c);
int ans=0;
for(int i=1;i<v.size();i++){
//cout<<v[i].x-v[i-1].x<<" "<<query(0,maxh,1)<<endl;
ans+=(v[i].x-v[i-1].x)*query(0,maxh,1);
insert(v[i].y1,v[i].y2,1,v[i].c);
}
cout<<ans<<endl;
}
int main(){
while(input()){
solve();
}
solve();
return 0;
}
最新文章
- javascript中的事件冒泡和事件捕获
- kettle etl
- 1458: 士兵占领 - BZOJ
- c语言数组的初始化
- iOS 音频开发经验汇总
- svn服务器配置小记
- Android数据的四种存储方式SharedPreferences、SQLite、Content Provider和File (三) —— SharePreferences
- C++可变参数的另一种实现
- Just4Fun - Comparaison between const and readonly in C#
- 使用(Drawable)资源——图片资源
- [SDOI2017]数字表格
- VIM编辑器操作命令积累
- centos7 与 archlinux用户 安装 python3模块 pytaglib
- 1、JUC--volatile 关键字-内存可见性
- IDEA git修改远程仓库地址
- Dropwizard简单入门
- 关于构造器和Serlvet的知识点
- 如何在 Ubuntu 中安装 QGit 客户端
- ECMall2.x模板制作入门系列之2(模板标签/语法)
- 转:Oracle下创建ASM磁盘总结