UVa 1382 Distant Galaxy 解题报告（枚举 + 前缀和）

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本文探讨如何通过观察远处星系并记录星系中星系系统的坐标，计算出包含最多星系系统的矩形的面积。通过输入星系数量及其坐标，输出最大矩形面积。此题旨在深入理解空间几何与星系分布的关系。

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1382 - Distant Galaxy

Time limit: 3.000 seconds

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

解题报告：绝对的好题目。

《训练指南》上说的很清楚了，不说啥了。总之，仔细体会，必有收获。代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <string>
using namespace std;

#define ff(i, n) for(int i=0;i<(n);i++)
#define fff(i, n, m) for(int i=(n);i<=(m);i++)
#define dff(i, n, m) for(int i=(n);i>=(m);i--)
#define mem(a) memset((a), 0, sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
void work();

int main()
{
#ifdef ACM
    freopen("in.txt", "r", stdin);
//    freopen("in.txt", "w", stdout);
#endif // ACM

    work();
}

/*****************************************/

struct Point
{
    int x, y;
    bool operator<(const Point & cmp) const
    {
        return x < cmp.x;
    }
} p[111];

int n, m, y[111], l[111], on[111], on2[111];

int solve()
{
    sort(p, p+n);
    sort(y, y+n);
    m = unique(y, y+n)-y;

    if(m <= 2) return n;

    int ans = 0;
    ff(a, m) fff(b, a+1, m)
    {
        int miny = y[a], maxy = y[b];

        int k = 0;
        ff(i, n)
        {
            if(i==0 || p[i].x != p[i-1].x)
            {
                k++;
                on[k] = on2[k] = 0;
                l[k] = l[k-1]+on2[k-1]-on[k-1];
            }

            if(p[i].y > miny && p[i].y < maxy) on[k]++;
            if(p[i].y >= miny && p[i].y <= maxy) on2[k]++;
        }

        int M = 0;
        fff(j, 1, k)
        {
            ans = max(ans, l[j]+on2[j]+M);
            M = max(M, on[j]-l[j]);
        }
    }

    return ans;
}

void work()
{
    int cas = 1;
    while(~scanf("%d", &n) && n)
    {
        ff(i, n)
        {
            scanf("%d%d", &p[i].x, &p[i].y);
            y[i] = p[i].y;
        }

        printf("Case %d: %d\n", cas++, solve());
    }
}