LA 3695 Distant Galaxy（扫描线）

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#扫瞄线

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本文探讨了一种算法，用于解决寻找遥远星系中包含最多星系边界的矩形问题。通过枚举上下边界并使用扫描线算法，该算法能够在一系列给定的星系坐标中找到最优解。

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Distant Galaxy

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVALive. Original ID: 3695
64-bit integer IO format: %lld Java class name: Main

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You are observing a distant galaxy using a telescope above the Astronomy Tower, and you think that a rectangle drawn in that galaxy whose edges are parallel to coordinate axes and contain maximum star systems on its edges has a great deal to do with the mysteries of universe. However you do not have the laptop with you, thus you have written the coordinates of all star systems down on a piece of paper and decide to work out the result later. Can you finish this task?

$\epsfbox{p3694.eps}$

Input

There are multiple test cases in the input file. Each test case starts with one integer N , (1 $\le$ N $\le$ 100) , the number of star systems on the telescope. N lines follow, each line consists of two integers: the X and Y coordinates of the K -th planet system. The absolute value of any coordinate is no more than 10⁹ , and you can assume that the planets are arbitrarily distributed in the universe.

N = 0 indicates the end of input file and should not be processed by your program.

Output

For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.

Sample Input

Sample Output

Case 1: 7

Source

Regionals 2006, Asia - Shanghai

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给出n个点坐标，找一个矩形，使边界上的点最多。思路枚举上下边界，然后用扫描线，大白P53

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 10;
struct Node {
    int x,y;
    bool operator < (const Node & rhs) const {
        return x < rhs.x;
    }
}p[maxn];
int y[maxn];
int L[maxn],on[maxn],on2[maxn];
int solve(int n)
{
    sort(p,p+n);
    sort(y,y+n);

    int m = unique(y,y+n) - y;
    if(m < 3) return n; ///不超过两个y坐标
    int ans = 0;
    for(int a = 0; a < m; ++a) {
        for(int b = a+1; b < m; ++b) {
            int k = 0,ymin = y[a],ymax = y[b];
            for(int i = 0; i < n; ++i) {
                if(i==0 || p[i].x != p[i-1].x) { ///新的竖线
                    ++k;
                    on[k] = 0;
                    on2[k] = 0;
                    L[k] = L[k-1] + on2[k-1] - on[k-1];
                }
                if(p[i].y > ymin && p[i].y < ymax) ++on[k];
                if(p[i].y >= ymin && p[i].y <= ymax) ++on2[k];
            }
            if(k < 3) return n; ///不超过两个x坐标
            int M = 0;
            for(int i = 1; i <= k; ++i) {
                ans = max(ans,L[i]+on2[i]+M);
                M = max(M,on[i]-L[i]);
            }
        }
    }
    return ans;
}
int main()
{
    int n,cas = 0;
    while(scanf("%d",&n)==1&& n) {
        ++cas;
        for(int i = 0; i < n; ++i) {
            scanf("%d%d",&p[i].x,&p[i].y);
            y[i] = p[i].y;
        }
        int ans = solve(n);
        printf("Case %d: %d\n",cas,ans);
    }
    return 0;
}