Description
Baker Crumble has just baked N triangular burek
2
pastries. Each pastry can be represented in the
Cartesian coordinate system as a triangle with vertices in integer coordinate points.
The baker's mischievous son Joey has just taken a large knife and started to cut the pastries. Each cut
that Joey makes corresponds to a horizontal (y = c) or vertical (x = c) line in the coordinate system.
Help the baker assess the damage caused by Joey's pastry cutting. Your task is to determine, for each
Joey's cut, how many pastries are affected (such that both the left and right parts of the cut pastry have
areas greater than zero).
Input
The first line of input contains the positive integer N (2 ≤ N ≤ 100 000), the number of burek pastries.
Each of the following N lines contains six nonnegative integers smaller than 10
6
. These numbers are, in
order, the coordinates (x1, y1), (x2, y2), (x3, y3) of the three pastry-triangle vertices. The three vertices will
not all be on the same line. The pastries can touch as well as overlap.
The following line contains the positive integer M (2 ≤ M ≤ 100 000), the number of cuts.
Each of the following M lines contains a single cut line equation: “x = c” or “y = c” (note the spaces
around the equals sign), where c is a nonnegative integer smaller than 10
6
.
Output
For each cut, output a line containing the required number of cut pastries.
Sample Input
3
1 0 0 2 2 2
1 3 3 5 4 0
5 4 4 5 4 4
4
x = 4
x = 1
y = 3
y = 1
4
2 7 6 0 0 5
7 1 7 10 11 11
5 10 2 9 6 8
1 9 10 10 4 1
4
y = 6
x = 2
x = 4
x = 9
Sample Output
0
1
1
2
3
2
3
2
HINT
In test data worth at least 40 points, M ≤ 300.
In test data worth an additional 40 points, the vertex coordinates of all triangles will be smaller than
1000.
分析：分兩邊處理，先從六個數找出x最大最小、y最大最小四個數，如為x=n軸時，只需搜索小於n的x最大的所有數a，大於n的x最小的所有數b，答案即是n-a-b，y軸時亦同樣分析。
代碼：
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAX=1000000;
int num[4][MAX+2];
int main()
{
int n,m,i,ans,in;
while(scanf("%d",&n)!=EOF)
{
memset(num,0,sizeof(num));
for(i=0;i<n;i++)
{
int x1,y1,x2,y2,x3,y3,max_x,max_y,min_x,min_y;
scanf("%d%d%d%d%d%d",&x1,&y1,&x2,&y2,&x3,&y3);
max_x=max(x1,max(x2,x3));
min_x=min(x1,min(x2,x3));
max_y=max(y1,max(y2,y3));
min_y=min(y1,min(y2,y3));
num[0][max_x]++;
num[1][min_x]++;
num[2][max_y]++;
num[3][min_y]++;
}
for(i=1;i<=MAX;i++)
{
num[0][i]+=num[0][i-1];
num[2][i]+=num[2][i-1];
}
for(i=MAX-1;i>=0;i--)
{
num[1][i]+=num[1][i+1];
num[3][i]+=num[3][i+1];
}
scanf("%d",&m);
while(m--)
{
char word;
scanf(" %c = %d",&word,&in);
if(word=='x')
ans=n-num[0][in]-num[1][in];
else
ans=n-num[2][in]-num[3][in];
printf("%d\n",ans);
}
}
return 0;
}
