【POJ】2002 - Squares（暴力枚举 & 双关键字二分查找）
【POJ】2002 - Squares（暴力枚举 & 双关键字二分查找）
题目链接：点击打开题目
Squares
Time Limit: 3500MS | Memory Limit: 65536K | |
|---|---|---|
Total Submissions: 19211 | Accepted: 7417 |
Description
A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property. So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.
Input
The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.
Output
For each test case, print on a line the number of squares one can form from the given stars.
Sample Input
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0Sample Output
1
6
1Source
Rocky Mountain 2004
简单粗暴，每次枚举两个点，以这两个点作为边判断另外两个点是否存在，判断的时候用二分查找。
代码如下：
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define CLR(a,b) memset(a,b,sizeof(a))
#define INF 0x7fffffff
#define LL long long
#define PI acos(-1.0)
struct node
{
int x,y;
}d[1011];
int n;
bool cmp(node a,node b)
{
if (a.x == b.x)
return a.y < b.y;
return a.x < b.x;
}
bool find(int x,int y) //二分找点
{
int l = 1;
int r = n;
while (r >= l)
{
int mid = (l + r) >> 1;
if (d[mid].x < x)
l = mid + 1;
else if (d[mid].x > x)
r = mid - 1;
else if (d[mid].y < y)
l = mid + 1;
else if (d[mid].y > y)
r = mid - 1;
else
return true; //找到了该点
}
return false;
}
int main()
{
while (~scanf ("%d",&n) && n)
{
for (int i = 1 ; i <= n ; i++)
scanf ("%d %d",&d[i].x,&d[i].y);
sort(d+1,d+1+n,cmp);
int ans = 0;
for (int i = 1 ; i < n ; i++)
{
for (int j = i + 1 ; j <= n ; j++)
{
int tx = d[i].x - d[j].x;
int ty = d[i].y - d[j].y;
if (find(d[i].x + ty , d[i].y - tx) && find(d[j].x + ty , d[j].y - tx))
ans++;
if (find(d[i].x - ty , d[i].y + tx) && find(d[j].x - ty , d[j].y + tx))
ans++;
}
}
printf ("%d\n",ans/4); //多算了3次，除以4
}
return 0;
}腾讯云开发者
