Calculate the number of toys that land in each bin of a partitioned toy box.?

計算每一個玩具箱里面玩具的數量

Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys.?
媽媽和爸爸有一個問題，他們的孩子約翰從來沒有在玩完玩具后把玩具都放好，他們有了約翰一個矩形的箱子用來放他的玩具，但是約翰很叛逆，他服從了他父母，不過只是簡單地把玩具扔進了箱子里。所有的玩具都搞混了，并且約翰不可能找到他最喜歡的玩具。

John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.?

約翰的父母想出了如下的注意，他們在箱子里放了紙板進行分區，即使約翰不停地把玩具扔進箱子里，至少玩具被扔進不同的垃圾箱并保持分開，下圖展現了樣例玩具箱的俯視圖。

For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.

對于這個問題，當約翰把玩具扔進箱子里時，你被要求確定有多少個玩具被扔進了分區。

Input

The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.

?????輸入文件包含一個或多個問題，第一行是第一個問題包含6個數，n,m,x1,y1,x2,y2,紙板分區的數目為n（0 < n <= 5000），玩具的數量是m（0＜m＜5000），箱子左上角的坐標為（x1,y1），右下角的坐標為（x2,y2）。以下N行，每行包含兩個整數，Ui Li，表示第i個紙板分區的結束是在坐標（UI，y1）和（Li，y2）。你可以假設紙板分區彼此不相交，它們是從左到右按排序順序指定的。接下來的m行每行包含兩個整數，Xj Yj指定該位置被第j個玩具落在了箱子里，玩具位置的順序是隨機的。你可以假設沒有玩具會準確地落在硬紙板的隔板上或盒子的外面（就是所有玩具肯定都會掉進箱子里面），輸入由一個單0組成的行終止。

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Output

The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.

每個問題的輸出將是玩具箱中每個獨立的箱子的一行。對于每一個箱子，打印它的箱子的號碼，后面是冒號和一個空格，后面跟著丟進箱子的玩具數量。箱編號從0（最左邊箱子）N（最右邊的箱子）。用一條空行分隔不同問題的輸出。

Sample?Input

5?6?0?10?60?0 3?1 4?3 6?8 10?10 15?30 1?5 2?1 2?8 5?5 40?10 7?9 4?10?0?10?100?0 20?20 40?40 60?60 80?80 5?10 15?10 25?10 35?10 45?10 55?10 65?10 75?10 85?10 95?10 0

Sample?Output

0:?2 1:?1 2:?1 3:?1 4:?0 5:?10:?2 1:?2 2:?2 3:?2 4:?2?

Hint

As the example illustrates, toys that fall on the boundary of the box are "in" the box.
如圖所示，落在盒子邊界上的玩具在盒子里

分析：

給n條線段，構成n+1個區間，然后是m個點的坐標，問0-n個區間，每個區間有多少個點？

我們先來看一下只有1條線段的情況，一條線段把平面分成兩個區間，如果點在線段的左側，那么這個點在左區間（0號），否則在右區間（1號）；

點(u,y1)和點(x,y)連線的斜率為k1=?，點(l,y2)和點(x,y)連線的斜率為k2=?

k1=k2,三點共線

k1>k2,點在直線的右側

k1<k2,點在直線的左側

POJ2318代碼：

import java.util.*; import java.math.*; class point{//記錄坐標int x;int y;public void point(int _x,int _y){this.x=_x;this.y=_y;} }class node{//隔板上下的點point a,b;public void node(point a,point b) {this.a=a;this.b=b;} }public class Main {static int maxn=(int)5e3+10;static int n,m,x1,y1,x2,y2;static node[] A=new node[maxn];static int[] ans=new int[maxn];//判斷點(x,y)與線段A[mid]的關系static boolean judge(int x,int y,int mid) {point a=A[mid].a; point b=A[mid].b;if((a.x-x)*(b.y-y)-(b.x-x)*(a.y-y)>0)return true;//點在線段右邊return false;//點在線段左邊}//二分查找點(x,y)最近的線段static void search(int x,int y,int n) {int left=0,right=n-1;while(left<=right) {int mid=(left+right)>>1;if(judge(x,y,mid)) //點在mid所在那條線段的右側left=mid+1;else //點在mid所在那條線段的左側right=mid-1;}ans[left]++;//記錄不同箱子所得到的玩具的數量}public static void main(String[] args) {Scanner cin=new Scanner(System.in);int ca=1;while(cin.hasNext()) {if(ca!=1) System.out.println();ca++;n=cin.nextInt();if(n==0) break;m=cin.nextInt();x1=cin.nextInt();y1=cin.nextInt();x2=cin.nextInt();y2=cin.nextInt();for(int i=0;i<n;i++) {int u,l;u=cin.nextInt();l=cin.nextInt();node tmp=new node();point a=new point();point b=new point();a.point(u, y1);b.point(l, y2);tmp.node(a, b);A[i]=tmp;}//存儲隔板的位置for(int i=0;i<=n;i++) {//初始化ans[i]=0;}for(int i=0;i<m;i++) {int x=cin.nextInt();int y=cin.nextInt();search(x,y,n);//尋找這個玩具在那個區間}for(int i=0;i<=n;i++) {System.out.println(i+": "+ans[i]);}}cin.close();} } #include<cstdio> #include<cstdlib> #include<cstring> using namespace std; struct point{int x,y; };//記錄坐標 struct Node{point a,b; }A[5010];//隔板上下的點 int pos[5010]; //判斷點(x,y)與線段A[mid]的關系 bool judge(int xx,int yy,int mid){int ans=(A[mid].a.x-xx)*(A[mid].b.y-yy)-(A[mid].a.y-yy)*(A[mid].b.x-xx);if(ans<0)return false;//點在線段左邊return true;//點在線段右邊 } //二分查找點(x,y)最近的線段 void search(int xx,int yy,int n){int left=0,right=n-1;while(left<=right){int mid=(left+right)>>1;if(judge(xx,yy,mid)){//點在mid所在那條線段的右側left=mid+1;}else { //點在mid所在那條線段的左側right=mid-1;}}pos[left]++;//記錄不同箱子所得到的玩具的數量 } int main() {int n,m,i,j,x1,x2,y1,y2;while(scanf("%d",&n),n){scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);for(i=0;i<n;++i){int xd,xu;scanf("%d%d",&xu,&xd);A[i].a.x=xu;A[i].a.y=y1;A[i].b.x=xd;A[i].b.y=y2;}//存儲隔板的位置memset(pos,0,sizeof(pos));for(i=0;i<m;++i){int xx,yy;scanf("%d%d",&xx,&yy);search(xx,yy,n);}//尋找這個玩具在那個區間for(i=0;i<=n;++i){printf("%d: %d\n",i,pos[i]);}printf("\n");}return 0; }

POJ 2398：先給N條線段從小到大排序，然后二分查找

import java.util.*; import java.math.*; class point{//記錄坐標int x;int y;public void point(int _x,int _y){this.x=_x;this.y=_y;} }class node{//隔板上下的點point a,b;public void node(point a,point b) {this.a=a;this.b=b;} }//比較器，x坐標從小到大排序；x相同時，按照y從小到大排序 /* class MyComparator implements Comparator{public int compare(Object arg0,Object arg1) {point a=(point)arg0;point b=(point)arg1;if(a.x!=b.x)return a.x>b.x?1:-1;return a.y>b.y?1:-1;} } */ class MyComparator implements Comparator<point>{public int compare(point a,point b) {if(a.x!=b.x)return a.x>b.x?1:-1;return a.y>b.y?1:-1;} } public class Main {static int maxn=(int)5e3+10;static int n,m,x1,y1,x2,y2;static node[] A=new node[maxn];static int[] ans=new int[maxn];//判斷點(x,y)與線段A[mid]的關系static boolean judge(int x,int y,int mid) {point a=A[mid].a; point b=A[mid].b;if((a.x-x)*(b.y-y)-(b.x-x)*(a.y-y)>0)return true;//點在線段右邊return false;//點在線段左邊}//二分查找點(x,y)最近的線段static void search(int x,int y,int n) {int left=0,right=n-1;while(left<=right) {int mid=(left+right)>>1;if(judge(x,y,mid)) //點在mid所在那條線段的右側left=mid+1;else //點在mid所在那條線段的左側right=mid-1;}ans[left]++;//記錄不同箱子所得到的玩具的數量}public static void main(String[] args) {Scanner cin=new Scanner(System.in);int ca=1;while(cin.hasNext()) {//if(ca!=1) System.out.println();ca++;n=cin.nextInt();if(n==0) break;m=cin.nextInt();x1=cin.nextInt();y1=cin.nextInt();x2=cin.nextInt();y2=cin.nextInt();//List<point> list=new ArrayList<point>();point[] arr=new point[n];for(int i=0;i<n;i++) {int u,l;u=cin.nextInt();l=cin.nextInt();point p=new point();p.point(u, l);//list.add(p);arr[i]=p;}//存儲隔板的位置Comparator cmp=new MyComparator();Arrays.sort(arr, cmp);//list.sort(cmp);for(int i=0;i<n;i++) {//初始化//point p=list.get(i);point p=arr[i];node tmp=new node();point a=new point();point b=new point();a.point(p.x, y1);b.point(p.y, y2);tmp.node(a, b);A[i]=tmp;ans[i]=0;}ans[n]=0;for(int i=0;i<m;i++) {int x=cin.nextInt();int y=cin.nextInt();search(x,y,n);//尋找這個玩具在那個區間}int[] cnt=new int[1010];for(int i=1;i<1010;i++)cnt[i]=0;int max=-1;for(int i=0;i<=n;i++) {if(ans[i]!=0) {cnt[ans[i]]++;max=ans[i]>max?ans[i]:max;}}System.out.println("Box");for(int i=1;i<=max;i++)if(cnt[i]!=0)System.out.println(i+": "+cnt[i]);}cin.close();} }

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