poj 2318
Total Submissions: 13667 | Accepted: 6596 |
Description
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
給你n條直線,這些直線把一個矩形分成了n+1個區域,給你一些點,統計每個區域點的個數
思路:枚舉每個點,然後判斷點在哪個區域,通過叉積的符號來判斷點在直線左側還是右側,可以用二分實現
#include<iostream>
#include<cstdio>
using namespace std;
struct Point
{
int x,y;
Point (int a=0,int b=0):x(a),y(b) {}
};
inline int multiply(Point sp,Point ep,Point op)
{
return((sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y));
}
Point a[5001],b[5001];
int c[5001];
int main()
{
int N,M,x1,x2,y1,y2,n,m,mid,i;
while(~scanf("%d",&N))
{
if(N<=0) break;
for(i=0; i<5001; i++)
c[i]=0;
scanf("%d %d %d %d %d",&M,&x1,&y1,&x2,&y2);
for(i=0; i<N; i++)
{
scanf("%d %d",&b[i].x,&b[i].y);
//(b[i].x,y1) (b[i].y,y2);
}
int s=0;
for(i=0; i<M; i++)//toys
scanf("%d %d",&a[i].x,&a[i].y);
for(i=0; i<M; i++)
{
int left=0,right=N-1;
while(left<=right)
{
mid=(left+right)/2;
if(multiply(Point(b[mid].x,y1),Point(b[mid].y,y2),a[i])<0)//叉積小於0說明點在直線左側
right=mid-1;
else left=mid+1;
}
c[left]++;
}
for(i=0; i<=N; i++)
printf("%d: %d\n",i,c[i]);
printf("\n");
}
return 0;
}