10574 - Counting Rectangles

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这篇博客讨论了一个关于在二维平面上确定正方形数量的问题。输入包括点的数量和它们的坐标，输出为找到的正方形总数。解决方案涉及坐标离散化和对角线枚举。

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Problem H

Counting Rectangles

Input: Standard Input

Output: Standard Output

Time Limit: 3 Seconds

Given n points on the XY plane, count how many regular rectangles are formed. A rectangle is regular if and only if its sides are all parallel to the axis.

Input

The first line contains the number of tests t(1<=t<=10). Each case contains a single line with a positive integer n(1<=n<=5000), the number of points. There are n lines follow, each line contains 2 integers x, y (0<=x, y<=10⁹) indicating the coordinates of a point.

Output

For each test case, print the case number and a single integer, the number of regular rectangles found.

Sample Input Output for Sample Input

0 0

2 0

0 2

2 2

1 1

0 0

0 30

0 900

Case 1: 1

Case 2: 0

Problemsetter: Rujia Liu, Member of Elite Problemsetters' Panel

Special Thanks: IOI2003 China National Training Team

思路：首先将x坐标和y坐标离散化一下....然后就开始直接乱搞了...有人n^2枚举对角线就可以了。我的麻烦一些咯。

代码：

#include <iostream>
#include <vector>
#include <algorithm>
#include <string.h>
#include <cstring>
#include <map>
#include <set>
#include <stdio.h>
#include <cmath>
#include <cassert>
#include <math.h>
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define rrep(i,b,a) for(int i = (b); i >= (a); --i)
#define clr(a,x) memset(a,(x),sizeof(a))
#define LL long long
#define eps 1e-9
#define mp make_pair
using namespace std;
const int maxn = 5000 + 5;
int n;
struct Point
{
    int x,y;
    bool operator < (const Point & pt) const
    {
        if (x != pt.x) return x < pt.x;
        return y < pt.y;
    }
}pt[maxn];

bool vis[maxn][maxn];

vector<int> line[maxn];

void input()
{
    scanf("%d",&n);
    rep(i,0,n) scanf("%d%d",&pt[i].x,&pt[i].y);
    sort(pt,pt+n);
    vector<int> v;
    rep(i,0,n) {
        v.push_back(pt[i].x);
    }
    sort(v.begin(),v.end());
    v.erase(unique(v.begin(),v.end()),v.end());
    rep(i,0,n) {
        pt[i].x = lower_bound(v.begin(),v.end(),pt[i].x) - v.begin();

    }
    v.clear();
    rep(i,0,n) v.push_back(pt[i].y);
    sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end());
    rep(i,0,n) pt[i].y = lower_bound(v.begin(),v.end(),pt[i].y) - v.begin();
    rep(i,0,n) vis[pt[i].x][pt[i].y] = true;

    rep(i,0,n+1) line[i].clear();
    rep(i,0,n) line[pt[i].x].push_back(i);
}

bool visy[maxn];
vector<int> vx[maxn];

void solve()
{
    int sz = 0;
    clr(visy,0);
    rep(i,0,maxn) vx[i].clear();
    rep(i,0,n) {
        if (line[i].empty()) break;
        sz++;
    }
    vector<int> * v;
    int ans = 0;
    int a,y1,b,y2;
    rrep(w,sz-1,0) {
        rep(i,0,line[w].size()) {
            a = line[w][i];
            y1 = pt[a].y;
            if (!visy[y1]) continue;
            rep(j,i+1,line[w].size()) {
                b = line[w][j];
                y2 = pt[b].y;
                if (!visy[y2]) continue;
                if (vx[y1].size() < vx[y2].size()) v = &vx[y1];
                else v = &vx[y2];
                for(auto it = v->begin(); it != v->end(); ++it) {
                    if (vis[*it][y1] && vis[*it][y2]) ++ans;
                }
            }
        }
        rep(i,0,line[w].size()) {
            int y = pt[line[w][i]].y;
            vx[y].push_back(w);
            visy[y] = true;
        }
    }
    printf("%d\n",ans);
    rep(i,0,n) {
        int x = pt[i].x, y = pt[i].y;
        vis[x][y] = false;
    }
}

void Getinput()
{
    freopen("in.txt","w",stdout);
    int T = 1;
    printf("%d\n",T);
    while (T--) {
        int n = 5000;
        printf("%d\n",n);
        rep(i,0,n/2) printf("%d %d\n",0,i);
        rep(i,0,n-n/2) printf("1 %d\n",i);
    }
}

int main()
{
   //Getinput(); return 0;
    #ifdef ACM
        freopen("in.txt", "r", stdin);
       // freopen("out.txt","w",stdout);
    #endif // ACM
    int T; cin >> T;
    rep(cas,1,T+1) {
        input();
        printf("Case %d: ",cas);
        solve();
    }
}