排座椅

题目地址：

https://ac.nowcoder.com/acm/problem/16618

基本思路：

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false); cin.tie(0)
#define int long long
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1010;
int n,m,k,l,d;
struct Node{
    int num,pos;
}x[maxn],y[maxn];
bool cmp1(Node a,Node b){
  return a.num > b.num;
}
bool cmp2(Node a,Node b){
  return a.pos < b.pos;
}
signed main() {
  IO;
  cin >> n >> m >> k >> l >> d;
  for(int i = 1 ; i <= d ; i++) {
    int x_1, y_1, x_2, y_2;
    cin >> x_1 >> y_1 >> x_2 >> y_2;
    if (x_1 == x_2) {
      y[min(y_1, y_2)].num++; // 这个位置隔开能减少的交头接耳人数;
      y[min(y_1, y_2)].pos = min(y_1, y_2); // 隔开的位置;
    } else {
      x[min(x_1, x_2)].num++;
      x[min(x_1, x_2)].pos = min(x_1, x_2);
    }
  }
  // 按照隔开能减少的交头接耳人数从大到小排序;
  sort(x + 1,x + n + 1,cmp1);
  sort(y + 1,y + m + 1,cmp1);

  // 将最优的前k行，前l列从小到大顺序排一下;
  sort(x + 1,x + k + 1,cmp2);
  sort(y + 1,y + l + 1,cmp2);

  rep(i,1,k) cout << x[i].pos << ' ';
  cout << '\n';

  rep(i,1,l) cout << y[i].pos << ' ';
  cout << '\n';
  return 0;
}