2018.07.03 POJ 2318 TOYS（二分+简单计算几何）

TOYS

Time Limit: 2000MS Memory Limit: 65536K

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

这是一道基础的计算几何，就是让你计算用线段分出的不规则图形的区域中分别有几个点。这题我们直接二分答案，如果点在当前第mid" role="presentation" style="position: relative;">midmid条线段的左边，那么我们移动右指针，否则我们移动左指针。怎么判断方向呢？直接使用向量的叉积来判断就可以了。

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 5005
using namespace std;
struct pot{
    int x,y;
};
struct line{pot a,b;}l[N];
int n,m,cnt[N];
int x1,x2,y1,y2;
inline int cross(pot a,pot b){return a.x*b.y-a.y*b.x;}
inline bool dir(int k,pot p){
    pot a,b;
    a.x=l[k].a.x-p.x,a.y=l[k].a.y-p.y;
    b.x=l[k].b.x-p.x,b.y=l[k].b.y-p.y;
    return cross(a,b)>0;
}
inline int search(pot p){
    int l=1,r=n,ret;
    while(l<=r){
        int mid=(l+r>>1);
        if(dir(mid,p))l=mid+1;
        else r=mid-1;
    }
    return r;
}
int main(){
    while(scanf("%d",&n)){
        if(n==0)break;
        memset(cnt,0,sizeof(cnt));
        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
        for(int i=1;i<=n;++i){
            l[i].a.y=y1,l[i].b.y=y2;
            scanf("%d%d",&l[i].a.x,&l[i].b.x);
        }
        for(int i=1;i<=m;++i){
            pot p;
            scanf("%d%d",&p.x,&p.y);
            ++cnt[search(p)];
        }
        for(int i=0;i<=n;++i)printf("%d: %d\n",i,cnt[i]);
        puts("");
    }
    return 0;
}

巴特西

2018.07.03 POJ 2318 TOYS（二分+简单计算几何）

最新文章

热门文章