TOYS POJ

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.

跟我最初的想法完全不同ORZ

---<><><><><><><><><>------------------------------------第一道计算几何-------------------------------------------------超级无敌分割线-------------------------------------------------<><><><><><>

一开始我想的是从给出的点开始水平沿着x轴线负方向引出一条射线，看这个射线穿过了几个分界线那么他就是在第几格。

然后WA，就是WA，可能是因为表达射线的时候用到了很小的负数，然后cross运算的时候可能会溢出，就跪了。

果断搜索题解，发现根本不用引出射线直接叉乘就行QAQ。然后就过了。题解上还有二分，我没用二分，如果一条线在一个点的右边的时候，后面的点一定也是不需要判断了。

因为边是顺序给你的，直接break就不会超时的。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int M=5e3+10;int n,m,cnt[M],lx,uy,rx,dy;struct point{    int x1,y1,x2,y2;} edge[M];bool judge(int x,int y,int x1,int y1){    return x*y1-y*x1>=0;}int main(){    memset(cnt,0,sizeof(cnt));    while(scanf("%d%d%d%d%d%d",&n,&m,&lx,&uy,&rx,&dy)!=EOF&&n)    {        for(int i=0; i<n; i++)        {            scanf("%d%d",&edge[i].x1,&edge[i].x2);            edge[i].y1=uy,edge[i].y2=dy;        }        for(int i=0; i<m; i++)        {            int x,y,sum=0;            scanf("%d%d",&x,&y);            for(int j=0; j<n; j++)            {                if(judge(edge[j].x2-edge[j].x1,edge[j].y2-edge[j].y1,x-edge[j].x1,y-edge[j].y1))                {                    sum++;                }                else                    break;//当前边不符合以后一定不符合            }            cnt[sum]++;        }        for(int i=0; i<=n; i++)            printf("%d: %d\n",i,cnt[i]),cnt[i]=0;            putchar('\n');    }}