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TOYSTime Limit:2000MSMemory Limit:65536KTotal Submissions:11078Accepted:5312
Description
Calculate the number of toys that land in each bin of a partitioned toy box.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.
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.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
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.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.Sample Input
5 6 0 10 60 03 14 36 810 1015 301 52 12 85 540 107 94 10 0 10 100 020 2040 4060 6080 80 5 1015 1025 1035 1045 1055 1065 1075 1085 1095 100
Sample Output
0: 21: 12: 13: 14: 05: 10: 21: 22: 23: 24: 2
Hint
As the example illustrates, toys that fall on the boundary of the box are "in" the box.Source
Rocky Mountain 2003 n块木板将一个盒子分成n+1部分，给你一些玩具的坐标，求这n+1部分中每一部分包含多少个玩具，POJ 2318 TOYS 叉积应用
，电脑资料
《POJ 2318 TOYS 叉积应用》(https://www.unjs.com)。给的木板的坐标是排好序的，所以可以二分时用叉积进行判断玩具的位置。double过，long long wa。
//468 KB 219 ms#include<stdio.h>#include<string.h>#include#define M 5007#define ll long longusing namespace std;double U[M],L[M];int sum[M],n;double y_1,y2;bool chaji(double x,double y,int mid)//叉积判断是否在mid左面{ double x_1=U[mid]-L[mid]; double x_2=x-L[mid]; double y_2=y-y2; return (x_1*y_2-x_2*y_1)>=0;}double binary_search(double x,double y)//二分{ int s=1,e=n,mid; while(s<=e) { mid=(s+e)>>1; if(chaji(x,y,mid))e=mid-1; else s=mid+1; } return s;}int main(){ double x1,y1,x2; int m; int flag=0; while(scanf("%d",&n),n) { scanf("%d%lf%lf%lf%lf",&m,&x1,&y1,&x2,&y2); y_1=y1-y2; memset(U,0,sizeof(U)); memset(L,0,sizeof(L)); memset(sum,0,sizeof(sum)); for(int i=1;i<=n;i++) scanf("%lf%lf",&U[i],&L[i]); U[0]=x1;L[0]=x1; double x,y; while(m--) { scanf("%lf%lf",&x,&y); int pos=binary_search(x,y); sum[pos-1]++; } if(flag){printf("\n");}flag=1; for(int i=0;i<=n;i++)printf("%d: %d\n",i,sum[i]); } return 0;}