LA 3695 Distant Galaxy

https://icpcarchive.ecs.baylor.edu/index.php?option=onlinejudge&page=show_problem&problem=1696

　　今天还做回以前做过的一道题。

　　这题主要是离散化，然后将点的个数转化为前缀和。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>

using namespace std;
typedef vector<int> VI;

#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)
#define INC(i, a, b) for (int i = (a); i <= (b); i++)
#define DEC(i, a, b) for (int i = (a); i >= (b); i--)
#define PB push_back
#define _clr(x) memset(x, 0, sizeof(x))
#define SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

const int N = 111;
VI X, Y;
map<int, int> NX, NY;
int x[N], y[N];
int mat[N][N], preSum[N][N];

void input(int n) {
    NX.clear();
    NY.clear();
    X.clear();
    Y.clear();
    REP(i, n) {
        cin >> x[i] >> y[i];
        X.PB(x[i]);
        Y.PB(y[i]);
    }
    sort(ALL(X));
    sort(ALL(Y));
    int t = unique(ALL(X)) - X.begin();
    while (SZ(X) > t) X.pop_back();
    REP(i, t) {
        NX[X[i]] = i + 1;
    }
    t = unique(ALL(Y)) - Y.begin();
    while (SZ(Y) > t) Y.pop_back();
    REP(i, t) {
        NY[Y[i]] = i + 1;
    }
    _clr(mat);
    REP(i, n) {
        mat[NX[x[i]]][NY[y[i]]] = 1;
    }
//    REP_1(i, SZ(X)) {
//        REP_1(j, SZ(Y)) {
//            cout << mat[i][j];
//        }
//        cout << endl;
//    }
}

int work() {
    int nx = SZ(X), ny = SZ(Y), mx = 0;
    if (nx <= 1 || ny <= 1) return max(SZ(X), SZ(Y));
    REP_1(i, nx) {
        REP_1(j, ny) {
            preSum[i][j] = preSum[i - 1][j] + mat[i][j];
        }
    }
//    REP_1(i, nx) {
//        REP_1(j, ny) {
//            cout << preSum[i][j];
//        }
//        cout << endl;
//    }
    int sum, mn;
    REP_1(x1, nx) {
        INC(x2, x1 + 1, nx) {
            sum = 0;
            mn = 999999;
            REP_1(y, ny) {
                mx = max(mx, sum + preSum[x2][y] - preSum[x1 - 1][y] - mn);
                mn = min(mn, sum - preSum[x2 - 1][y] + preSum[x1][y]);
                sum += mat[x1][y] + mat[x2][y];
            }
        }
    }
    return mx;
}

int main() {
//    freopen("in", "r", stdin);
    int n, cas = 1;
    while (cin >> n && n) {
        input(n);
        cout << "Case " << cas++ << ": " << work() << endl;
    }
    return 0;
}

——written by Lyon