poj2318

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 8292		Accepted: 3929

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

题目大意：将m个玩具扔进一个从左到右分成n个块的箱子中，求每个分块里有多少个玩具，玩具不会在边界上（箱子的左上角坐标为(x1, y1)，箱子右下角坐标为(x2, y2)，中间n条分隔栏的上坐标的横坐标为U[i]，下坐标的横坐标为L[i]）。

第1次做计算几何题目，什么都不懂。看了别人的讲解才知道 叉积可以这样用。

向量的叉积：a = (x1,y1),b = (x2,y2), 则a x b = x1*y2 - x2*y1.
右手法则：若b在a的顺时针转动方向（夹角<90度），则a x b < 0,否则若b在a的逆时针方向,则a x b >0, 若a,b共线，a x b = 0.

看到数据比较大5000*5000，直接搜，感觉要超时，没想到658MS AC了，大概题目数据比较水吧。

后来看到别人的思路，才知道可以用二分，自己试了下 果然快很多180MS就过了。

自己为什么没有想到呢。唉

#include"stdio.h"
#include"string.h"
#define MAXN 5005
int i,j,n,m,x1,y1,x2,y2,t;
int ui[MAXN],li[MAXN],x,y,g[5005];
int find(int x1,int x2)
{
 int f1=(li[x1]-x)*(y1-y)-(ui[x1]-x)*(y2-y);
 int f2=(li[x2]-x)*(y1-y)-(ui[x2]-x)*(y2-y);
 if(f1<0&&f2>0)
  return 0;
 else
 if(f1>0)
  return -1;
 else
 if(f2<0)
  return 1;
}
void solve(int l,int r)
{
 t=(l+r)/2;
 int tt=find(t,t+1);
 if(tt==0)
  return;
 else if(tt==-1)
  solve(0,t);
 else
  solve(t+1,r);
}
int main()
{
 while(1)
 {
  scanf("%d",&n);
  if(n==0)break;
  scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);
  for(i=1;i<=n;i++)
  {
  
   scanf("%d%d",&ui[i],&li[i]);
  }
  ui[0]=x1,li[0]=x1;
  ui[n+1]=x2,li[n+1]=x2;
  memset(g,0,sizeof(g));
  for(i=0;i<m;i++)
  {
   scanf("%d %d",&x,&y);
   solve(0,n+1);
   g[t]++;
  }
  for(i=0;i<=n;i++)
   printf("%d: %d\n",i,g[i]);
  printf("\n");
 }
 return 0;
}