题目描述

参考zzugzx题解

Solution

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(__vv__) (__vv__).begin(), (__vv__).end()
#define endl "\n"
#define pai pair<int, int>
#define mk(__x__,__y__) make_pair(__x__,__y__)
#define ms(__x__,__val__) memset(__x__, __val__, sizeof(__x__))
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void print(ll x, int op = 10) { if (!x) { putchar('0'); if (op)    putchar(op); return; }    char F[40]; ll tmp = x > 0 ? x : -x;    if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]);    if (op)    putchar(op); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }
const int dir[][2] = { {0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1} };
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;

const int N = 2000 + 7;
int x[N], y[N], id[N];
vector<int> ans;

bool cmpx(int i, int j) {
    return x[i] > x[j];
}

bool cmpy(int i, int j) {
    return y[i] > y[j];
}

int main() {
    int n = read(), m = read(), k = read(), l = read(), d = read();
    for (int i = 1; i <= d; ++i) {
        int x1 = read(), y1 = read(), x2 = read(), y2 = read();
        if (y1 == y2)
            ++x[min(x1, x2)];
        else
            ++y[min(y1, y2)];
    }
    for (int i = 1; i <= n; ++i)    id[i] = i;
    sort(id + 1, id + 1 + n, cmpx);
    ans.clear();
    for (int i = 1; i <= k; ++i)
        ans.push_back(id[i]);
    sort(all(ans));
    for (auto it : ans)
        print(it, 32);
    puts("");
    for (int i = 1; i <= m; ++i)    id[i] = i;
    sort(id + 1, id + 1 + m, cmpy);
    ans.clear();
    for (int i = 1; i <= l; ++i)
        ans.push_back(id[i]);
    sort(all(ans));
    for (auto it : ans)
        print(it, 32);
    return 0;
}