POJ2318(KB13-A 计算几何)
TOYS
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 16222 | Accepted: 7779 |
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
Source
//2017-08-30
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; const int N = ; struct Point{
int x, y;
Point(){}
Point(int _x, int _y):x(_x), y(_y){}
//a-b 表示向量 ba
Point operator- (const Point &b) const {
return Point(x-b.x, y-b.y);
}
//向量叉积
int operator* (const Point &b) const {
return x*b.y - y*b.x;
}
}A, B; int ans[N], U[N], L[N];
int n, m; bool check(int id, int x, int y){
Point a(L[id], B.y);
Point b(U[id], A.y);
Point c(x, y);
//令I = 向量ab 叉乘 向量 bc,若I为正,点c在向量ab的左侧(沿向量方向看);为负则在右侧
return ((c-a)*(b-a)) > ;
} int get_position(int x, int y){
int l = , r = n+, mid, ans;
while(l <= r){
mid = (l+r)>>;
if(check(mid, x, y)){
ans = mid;
l = mid+;
}else r = mid-;
}
return ans;
} int main()
{
std::ios::sync_with_stdio(false);
//freopen("inputA.txt", "r", stdin);
while(cin>>n && n){
cin>>m>>A.x>>A.y>>B.x>>B.y;
U[] = L[] = A.x;
U[n+] = L[n+] = B.x;
for(int i = ; i <= n; i++)
cin>>U[i]>>L[i];
memset(ans, , sizeof(ans));
int x, y;
for(int i = ; i < m; i++){
cin>>x>>y;
ans[get_position(x, y)]++;
}
for(int i = ; i <= n; i++)
cout<<i<<": "<<ans[i]<<endl;
cout<<endl;
} return ;
}
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