POJ 2002.Squares

题目:http://poj.org/problem?id=2002

AC代码(C++):

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <vector>
#include <queue>
#include <math.h>
#include <string>
#include <string.h>
#include <bitset>

#define INF 0xfffffff
#define MAXN 100005

using namespace std;

const int N = 1005;
const int H = 997;

struct Node
{
int x,y;
int next;
};
Node node[N];
int cur;
int hashTable[H];
int n;

void initHash()
{
cur = 0;
for (int i = 0; i < H; ++i) hashTable[i] = -1;
}

void insertHash(int x, int y)
{
int h = abs(x+y)%H;
node[cur].x = x;
node[cur].y = y;
node[cur].next = hashTable[h];
hashTable[h] = cur;
cur++;
}

bool searchHash(int x, int y)
{
int h = abs(x+y)%H;
int next = hashTable[h];
while (next != -1)
{
if (node[next].x == x&&node[next].y == y) return true;
next = node[next].next;
}
return false;
}

int main(){
while(cin>>n){
if(n==0)break;
int ans = 0;
initHash();
for(int i = 0; i < n; i++){
int tmpx, tmpy;
cin>>tmpx>>tmpy;
insertHash(tmpx,tmpy);
}
for(int i = 0; i < cur; i++){
for(int j = 0; j < cur; j++){
if(i==j)continue;
int x3,y3,x4,y4;
x3 = node[j].y - node[i].y + node[j].x;
y3 = node[i].x - node[j].x + node[j].y;
x4 = node[j].y - node[i].y + node[i].x;
y4 = node[i].x - node[j].x + node[i].y;
if(searchHash(x3,y3)&&searchHash(x4,y4))ans++;
}
}
cout<<ans/4<<endl;
}
}

总结: 又是一道经典的哈希题. 总结下这一类的方法: 遇到像这样看似可以简单的枚举但一定会超时的题目, 就用哈希. 具体做法是, 将输入的对象拆分, 比如poj1840题(http://blog.csdn.net/njyexiong/article/details/78481702), 就是把等式拆分成两部分, 一边插入一边搜索. 这题也一样, 将正方形的四个点拆分成两个一对, 枚举输入的所有点对, 通过已知的两点求正方形中其他两个点的位置, 然后在哈希表中看这两个点是否存在, 存在则正方形存在.

关于已知正方形中相邻的两点(x1,y1), (x2,y2), 求其余两点的表达式:

x3 = y2 - y1 + x2

y3 = x1 - x2 + y2

x4 = y2 - y1 + x1

y4 = x1 - x2 + y1

因为一个正方形有4个边, 故一个正方形会重复计算4次, 所以最终答案应该/4.