poj2318——TOYS(计算几何+点与线段的位置)

2016-11-29 08:15:12来源:CSDN作者:blue_skyrim人点击

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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.
这里写图片描述
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

其实就是判断给出的m个点在n条线段与长方形组成图形的哪个区域。
如果在第n-1条边的右边且在第n条边的左边,那么这个点就在第n个区域,最后求n个区域拥有点的个数。
将这个点与线段的端点相连,求这两条边的叉积就能判断点在线段的那一边,线段的枚举可以由二分判断。
一开始脑补二分策略时判断反了,而通过看图又正确了,看来我二分学的不够好。

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 5010#define Mod 10001using namespace std;struct point{    int x,y;    point(){}    point(int a,int b)    {        x=a;        y=b;    }};struct Line{    point s,e;    Line(){}    Line(point a,point b)    {        s=a;        e=b;    }};int multy(point p0,point p1,point p2) //p0p1 x p0p2{    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}Line line[MAXN];int ans[MAXN];int main(){    int n,m,x1,y1,x2,y2;    bool ff=true;    while(~scanf("%d",&n)&&n)    {        if(ff)            ff=false;        else            printf("/n");        scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);        int u,l;        for(int i=0;i<n;++i)        {            scanf("%d%d",&u,&l);            line[i]=Line(point(u,y1),point(l,y2));        }        line[n]=Line(point(x2,y1),point(x2,y2)); //长方形的最右边也要参与判断        memset(ans,0,sizeof(ans));        point p;        int x,y;        for(int i=0;i<m;++i)        {            int low=0,high=n;            scanf("%d%d",&x,&y);            p=point(x,y);            int tmp;            while(low<=high)            {                int mid=(low+high)/2;                if(multy(p,line[mid].s,line[mid].e)<0)                {                    tmp=mid;                    high=mid-1;                }                else                    low=mid+1;            }            ans[tmp]++;        }        for(int i=0;i<=n;++i)            printf("%d: %d/n",i,ans[i]);    }    return 0;}
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