poj2398+poj2318(计算几何+二分)

Toy Storage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3461 Accepted: 2022

Description

Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:

We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.

Input

The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.

A line consisting of a single 0 terminates the input.

Output

For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 020 2080 8060 6040 405 1015 1095 1025 1065 1075 1035 1045 1055 1085 105 6 0 10 60 04 315 303 16 810 102 12 81 55 540 107 90

Sample Output

Box2: 5Box1: 42: 1

Source

Tehran 2003 Preliminary

      本题就是要判断点落在正方形的那个区域内。由于要在给定的时间复杂度范围内解决问题，所以题目说线段不想交且点不在线段上。所以可以直接差乘，二分求得点右边最靠近的线段。注意区域点的个数映射到拥有相同个数点的区域数。

       点与线段位置关系+二分

   

#include<iostream>#include<algorithm>using namespace std;const int MAXN=1000+100;int belong[MAXN];int  ans[MAXN];struct point{int x,y;point(){};point(int _x,int _y){x=_x;y=_y;}};struct line{point st,en;line(){}line(point _st,point _en){st=_st;en=_en;}}MyLine[MAXN];bool cmp(line a,line b){return a.st.x<b.st.x;}bool IsLeft(line a,point b){if((a.st.x-b.x)*(a.en.y-b.y)-(a.en.x-b.x)*(a.st.y-b.y)<0)return true;return false;}int main (){int n,m,x1,x2,y1,y2,Ui,Li;while(scanf("%d",&n),n){scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);for(int i=0;i<n;i++){scanf("%d%d",&Ui,&Li);MyLine[i]=line(point(Ui,y1),point(Li,y2));}MyLine[n]=line(point(x2,y1),point(x2,y2));sort(MyLine,MyLine+n+1,cmp);memset(belong,0,sizeof(belong));int tx,ty;while(m--){scanf("%d%d",&tx,&ty);point MyPoint(tx,ty);int low=0,high=n+1,mid;while(low<=high){mid=(low+high)>>1;if(IsLeft(MyLine[mid],MyPoint))high=mid-1;else low=mid+1;}belong[low]++;}memset(ans,0,sizeof(ans));for(int i=0;i<n+1;i++){if(belong[i])ans[belong[i]]++;}printf("Box\n");for(int i=0;i<n+1;i++){if(ans[i])printf("%d: %d\n",i,ans[i]);}}system("pause");return 0;}

     2138同上题。贴个代码。

 

#include<iostream>#include<algorithm>using namespace std;const int MAXN=5000+100;int belong[MAXN];int  ans[MAXN];struct point{int x,y;point(){};point(int _x,int _y){x=_x;y=_y;}};struct line{point st,en;line(){}line(point _st,point _en){st=_st;en=_en;}}MyLine[MAXN];bool cmp(line a,line b){return a.st.x<b.st.x;}bool IsLeft(line a,point b){if((a.st.x-b.x)*(a.en.y-b.y)-(a.en.x-b.x)*(a.st.y-b.y)<0)return true;return false;}int main (){int n,m,x1,x2,y1,y2,Ui,Li;while(scanf("%d",&n),n){scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);for(int i=0;i<n;i++){scanf("%d%d",&Ui,&Li);MyLine[i]=line(point(Ui,y1),point(Li,y2));}MyLine[n]=line(point(x2,y1),point(x2,y2));sort(MyLine,MyLine+n+1,cmp);memset(belong,0,sizeof(belong));int tx,ty;while(m--){scanf("%d%d",&tx,&ty);point MyPoint(tx,ty);int low=0,high=n+1,mid;while(low<=high){mid=(low+high)>>1;if(IsLeft(MyLine[mid],MyPoint))high=mid-1;else low=mid+1;}belong[low]++;}for(int i=0;i<n+1;i++){printf("%d: %d\n",i,belong[i]);}printf("\n");}system("pause");return 0;}