poj 2318 TOYS （点与线段位置关系判断）

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

Output

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

题目很简单，判断一下点与线段关系，枚举线段时，可以按区间二分枚举，左右边界。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
using namespace std;
#define N 5005
#define ll long long
int n,cnt[N];
struct p
{
    int x,y;
}a,b,c,up[N],down[N];
double cross(p a,p b,p c)
{
    return 1.0*(b.x-a.x)*(c.y-a.y)-1.0*(c.x-a.x)*(b.y-a.y);
}
int judge(p a)
{
    int l=1,r=n,mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        double s1=cross(a,down[mid-1],up[mid-1]);
        double s2=cross(a,down[mid],up[mid]);
        //printf("%d %.3f %.3f\n",mid,s1,s2);
        if(s1*s2<0)
            return mid;
        if(s2<0)
            l=mid+1;
        else
            r=mid-1;
    }
    return l;
}
int main()
{
    int i,t,m;
    while(scanf("%d",&n),n)
    {
        scanf("%d%d%d%d%d",&m,&b.x,&b.y,&c.x,&c.y);
        up[0].x=down[0].x=b.x;  //box起始点
        up[0].y=up[n+1].y=b.y;
        up[n+1].x=down[n+1].x=c.x; //box终点
        down[0].y=down[n+1].y=c.y;
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&up[i].x,&down[i].x);
            up[i].y=b.y;
            down[i].y=c.y;
        }
        n++;
        memset(cnt,0,sizeof(cnt));
        while(m--)
        {
            scanf("%d%d",&a.x,&a.y);
            t=judge(a);     // 二分判断位置关系
            cnt[t]++;
        }
        for(i=0;i<n;i++)
        {
            printf("%d: %d\n",i,cnt[i+1]);
        }
        puts("");
    }
    return 0;
}

poj 2318 TOYS （点与线段位置关系判断）

原文：http://blog.csdn.net/u011721440/article/details/41774839