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
题意：给出一个盒子以及N条线，把这个格子划分成N+1部分。然后给出一些坐标，统计这些坐标落在各个格子的次数。
线性规划，左上顶点一定是在每根线的左手边。这样就可以从左到右算出每根线的方程后，拿顶点坐标和当前坐标代入后计算其正负性，第一个同号的时刻就是他所在的格子。
#include <stdio.h>
#include <string.h>
int n,m,x1,y1,x2,y2,x,y;
int loc[5000][2];
double a[5000],b[5000];
int num[5001];
int i,j;
double v1,v2;
bool pd;
int main()
{
while (scanf("%d",&n)&&n)
{
if (m!=0) printf("\n");
memset(num,0,sizeof(num));
scanf ("%d %d %d %d %d",&m,&x1,&y1,&x2,&y2);
for (i=0;i<n;i++)
{
scanf("%d %d",&loc[i][0],&loc[i][1]);
a[i]=double(loc[i][1]-loc[i][0])/double(y1-y2);
b[i]=-a[i]*y1-loc[i][0];
}
for (i=0;i<m;i++)
{
scanf("%d %d",&x,&y);
pd=false;
for (j=0;j<n;j++)
{
if (x>loc[j][0]&&x>loc[j][1]) continue;
v1=x1+a[j]*y1+b[j];
v2=x+a[j]*y+b[j];
if (v1*v2>0) {
pd=true;
num[j]++;
break;
}
}
if (!pd) num[n]++;
}
for (i=0;i<=n;i++)
printf("%d: %d\n",i,num[i]);
}
return 0;
}