UVA - 1312 - Cricket Field（离散化）

题意：给定h * w（1 <= h, w <= 10^4）的区域，区域内的坐标范围为0~h和0~w，然后给定区域内n个点（0 <= n <= 100），求在区域内一个不包含点的最大面积的正方形（但点可以在正方形边界上），输出左下角坐标及边长（多解则任意输出）。

1、找出在h方向的所有边界；

2、二重循环枚举任意两个边界；

3、枚举范围内的点，构造矩形，对于每个矩形来说，长和宽其中较小的一个即为正方形的边长。

以上不会丢失最优解，即使有的矩形在h方向上可以扩展，但是枚举了任意h方向的两个边界，一定会取到。

以下是图示：

 

#include<cstdio>
#include<cstring>
#include<cctype>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#define fin freopen("in.txt", "r", stdin)
#define fout freopen("out.txt", "w", stdout)
#define pr(x) cout << #x << " : " << x << "   "
#define prln(x) cout << #x << " : " << x << endl
#define Min(a, b) a < b ? a : b
#define Max(a, b) a < b ? b : a
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const double pi = acos(-1.0);
const double EPS = 1e-6;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const ll MOD = 1e9 + 7;
using namespace std;

#define NDEBUG
#include<cassert>
const int MAXN = 100 + 10;
const int MAXT = 10000 + 10;

int T, h, w, n, y[MAXN], m;

struct Node{
    int x, y;
    bool operator < (const Node &rhs) const{
        return x < rhs.x || (x == rhs.x && y < rhs.y);
    }
}p[MAXN];

int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d%d%d", &n, &w, &h);
        for(int i = 0; i < n; ++i){
            scanf("%d%d", &p[i].x, &p[i].y);
            y[i] = p[i].y;
        }
        sort(p, p + n);
        y[n] = 0;  y[n + 1] = h;
        sort(y, y + n + 2);
        m = unique(y, y + n + 2) - y;
        int ans = 0, ansx = 0, ansy = 0;
        for(int i = 0; i < m; ++i)
            for(int j = i + 1; j < m; ++j){
                int miny = y[i], maxy = y[j], lur = 0;
                int ww = maxy - miny;
                for(int k = 0; k < n; ++k){
                    if(p[k].y >= maxy || p[k].y <= miny)  continue;
                    int hh = p[k].x - lur;
                    int tmp = Min(ww, hh);
                    if(tmp > ans){
                        ans = tmp;
                        ansx = lur;
                        ansy = miny;
                    }
                    lur = p[k].x;
                }
                int hh = w - lur;
                int tmp = Min(ww, hh);
                if(tmp > ans){
                    ans = tmp;
                    ansx = lur;
                    ansy = miny;
                }
            }
        printf("%d %d %d\n", ansx, ansy, ans);
        if(T)  printf("\n");
    }
    return 0;
}