# POJ 2318--TOYS(二分找点，叉积判断方向)

TOYS Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17974 Accepted: 8539 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 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: 1

0: 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. Source

Rocky Mountain 2003

• 题意:给n条边，划分成n+1个区域，再给定m个点坐标，点不会落在边界上和区域外，问每个区域中各自存在多少个点
• 代码如下
```#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
const int MAX = 5005;

typedef struct point {
int x;
int y;
}point;
typedef struct value {
point start;
point end;
}v;
v edge[MAX];
int sum[MAX];
int n, m, x1, y11, x2, y2, flag = 1;
point tp;
int Xj, Yj;
int multi(point p1, point p2, point p0) {  //判断p1p0和p2p0的关系，<0,p1p0在p2p0的逆时针方向
return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
}
void inset(point p) {
int low = 0, high = n;
while (low <= high) {
int mid = (high + low) / 2;
if (multi(p, edge[mid].start, edge[mid].end) < 0)    /*点p1在边的左侧*/
high = mid - 1;
else    //点p在边的右侧
low = mid + 1;
}
if (multi(p, edge[low-1].start, edge[low-1].end) < 0 )
sum[low-1]++;
else
sum[low]++;
}
int main() {
while (~scanf("%d", &n) && n) {
memset(sum, 0, sizeof(sum));
if (flag == 1)flag++;
else printf("\n");
scanf("%d%d%d%d%d", &m, &x1, &y11, &x2, &y2);
int Ui, Li;
for (int i = 0; i < n; i++) {
scanf("%d%d", &Ui, &Li);
edge[i].start.x = Ui;
edge[i].start.y = y11;
edge[i].end.x = Li;
edge[i].end.y = y2;
}
edge[n].start.x = x2;
edge[n].start.y = y11;
edge[n].end.x = x2;
edge[n].end.y = y2;
for (int j = 0; j < m; j++) {
scanf("%d%d", &Xj, &Yj);
tp.x = Xj;
tp.y = Yj;
inset(tp);
}
for (int i = 0; i <= n; i++)
printf("%d: %d\n", i, sum[i]);
}
return 0;
}```
• Experience:　前面点的构造写成

-

```1 edge[i].start = { Ui,y11 };
2 edge[i].end = { Li,y2 };

0 条评论

• ### POJ 2398--Toy Storage(叉积判断，二分找点，点排序)

Toy Storage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: ...

• ### POJ 2208--Pyramids(欧拉四面体体积计算)

Pyramids Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 344...

• ### 如何在Ubuntu 16.04上安装和配置Redis集群

Redis集群已经发展成为缓存，队列等的流行工具，因为它具有可扩展性和速度的潜力。本指南旨在使用三个Linode创建一个集群来演示分片。然后，如果发生故障，您将...

• ### 如何部署 Redis 集群

Redis 是我们目前大规模使用的缓存中间件，由于它强大高效而又便捷的功能，得到了广泛的使用。单节点的Redis已经就达到了很高的性能，为了提高可用性我们可以使...

• ### Salesforce平台支持多租户Multi tenant的核心设计思路

版权声明：本文为博主汪子熙原创文章，未经博主允许不得转载。 https://jerry.blog....

• ### Salesforce平台支持多租户Multi tenant的核心设计思路

Multitenancy is the fundamental technology that clouds use to share IT resources...

• ### 优化算法:到底是数学还是代码？

背景：我的一位同事曾提到，他在面试深度学习相关职位中被问到一些关于优化算法的问题。我决定在本文中就优化算法做一个简短的介绍。 成本函数的最优化算法 目标函数是...

• ### 目标检测一波接着一波 | YOLOv5又悄悄来袭！（附源码论文链接）

期待已久的检测经典又来来了一波强袭——yolov5。其实yolov5没有完整的文件，现在最重要的应该是把yolov4弄清楚，在目标检测领域中受益匪浅，可以在某些...