POJ 2318 叉积判断点与直线位置

时间：2016-03-06 17:14:50 阅读：226 评论：0 收藏：0 [点我收藏+]

TOYS

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.
技术分享

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

Hint

As the example illustrates, toys that fall on the boundary of the box are "in" the box.

题意：

一个矩形箱子，左上角与右下角的坐标给出，里面有n块板把箱子里的空间分隔成许多个分区，给出这些板在上边的x坐标、下边的x坐标，以及一堆玩具的坐标，求这些分区里的玩具数目。

题解：

记玩具在点p0，某块板的上边点是p1，下边点是p2，p2p1(向量)×p2p0>0表示p0在p1p2的左面，<0表示在右面。接下来就是用二分法找出每个点所在的分区。

叉积+二分查找

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1e6+10, M = 30005, mod = 1e9 + 7, inf = 0x3f3f3f3f;
typedef long long ll;
int n,m,x1,x2,y11,y2,ans[N],t1,t2;
struct point{int x,y;};
struct segment{point a,b;}s[N];


point sub(point a,point b) {//向量
    point t;
    t.x = a.x-b.x;
    t.y = a.y-b.y;
    return t;
}
int cross(point a,point b){//叉积公式
    return a.x*b.y-b.x*a.y;
}
int turn(point p1,point p2,point p3){ //叉积
    return cross(sub(p2,p1),sub(p3,p1));
}
void searchs(point x) {
    int l=1,r=n,mid,t=0;
    while(l<=r) {
        mid = (l+r)>>1;
        if(turn(s[mid].a,s[mid].b,x) >= 0) {
            t = mid;l=mid+1;
        }
        else r = mid-1;
    }
    ans[t]++;
}
int main() {
   while(scanf("%d",&n)&&n) {
        memset(ans,0,sizeof(ans));
    scanf("%d%d%d%d%d",&m,&x1,&y11,&x2,&y2);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&t1,&t2);
        s[i].a.x=t1;s[i].a.y=y11;
        s[i].b.x=t2;s[i].b.y=y2;
    }
    for(int i=1;i<=m;i++) {
        point t;
        scanf("%d%d",&t.x,&t.y);
        searchs(t);
    }
    for(int i=0;i<=n;i++)
        printf("%d: %d\n",i,ans[i]);
    printf("\n");
   }
}

POJ 2318 叉积判断点与直线位置

原文：http://www.cnblogs.com/zxhl/p/5247611.html

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