Coconuts hud5925 搜索+离散化

Coconuts

离散化

离散化判断联通块的个数还可以，但是这题竟然还让输出每个连通块的数量，除了签到题，其它题都太难了吧。
离散化墙的位置。
离散x坐标时，在离散x0 时，还要加入 x0 - 1该点也要进行离散化，才能确保两个不相邻的两点在离散化后仍然不相邻。
同时，每个点离散化后的坐标应该有一个权值表示空白快的宽度即可。

搜索

一块的面积即空白网格的数量。
广搜即可。

/*

hello world!

Just do it!

start time:2022-05-02 08:47:59.118827

*/

// #pragma GCC optimize (2)
// #pragma G++ optimize (2)
#include <bits/stdc++.h>
#define zero(x) memset(x, 0, sizeof(x));
#define one(x) memset(x, -1, sizeof(x));
#define m_INF(x) memset(x, 0x3f, sizeof(x));
#define m_inf(x) memset(x, 0x3f, sizeof(x));
#define m_f_INF(x) memset(x, -0x3f, sizeof(x));
#define m_f_inf(x) memset(x, -0x3f, sizeof(x));
#define all(x) x.begin(), x.end()
#define endl "\n" 
#define fi first
#define se second
#define lbt(x) ((x)&(-x))
#define pb push_back

struct cmpl{ template <typename A, typename B> bool operator()(const A &a1, const B &b1) { return b1 < a1; } };
struct cmpg{ template <typename A, typename B> bool operator()(const A &a1, const B &b1) { return a1 < b1; } };
#define p_ql(x) priority_queue<x, vector<x>, cmpl>
#define p_qlp(x, y) priority_queue<pair<x, y>, vector<pair<x, y>>, cmpl>
#define p_qg(x) priority_queue<x, vector<x>, cmpg>
#define p_qgp(x, y) priority_queue<pair<x, y>, vector<pair<x, y>>, cmpg>
template<class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
#define fo(i,from,to) for(int i=(from),ooo=(from)<(to)?1:-1,oooo=(to)+ooo;i!=oooo;i+=ooo)
#define fol(i,from,to) for(long long i=(from),ooo=(from)<(to)?1:-1,oooo=(to)+ooo;i!=oooo;i+=ooo)
#define foo(i,ooo) for(auto i=ooo.rbegin();i!=ooo.rend();++i)
#define foa(i,from,to) for(int i=(from),ooo=(to);i<=ooo;++i)
#define fos(i,from,to) for(int i=(from),ooo=(to);i>=ooo;--i)

using namespace std;

#ifndef LOCAL
#    define dbg(...) ;
#endif

#define itn int
#define int long long

#ifdef int
#define inf (0x3f3f3f3f3f3f3f3f)
#else
#define inf (0x3f3f3f3f)
#endif

typedef long long ll; typedef unsigned long long ull; typedef pair<int,int> pii;
const int  dir[8][2]={{0,1},{1,0},{0,-1},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}}, INF = 0x3f3f3f3f, f_inf = -1044266559, f_INF = -1044266559;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
const int N = 1e6 + 10;
int n, m;

void init()
{
    
}
pii a[210];
int p[500][500], rr, cc;
void solve()
{
    cin >> rr >> cc >> n;
	vector<int> vx = {0, rr, rr + 1}, vy = {0, cc, cc + 1};
	foa(i, 1, n) {
		int x, y;
		cin >> x >> y;
		a[i] = {x, y};
		vx.pb(x);
		vx.pb(x - 1);
		vy.pb(y);
		vy.pb(y - 1);
	}
	sort(all(vx));
	vx.erase(unique(all(vx)), vx.end());
	sort(all(vy));
	vy.erase(unique(all(vy)), vy.end());
	zero(p);
	map<int, int> mx, my;
	int r = vx.size() - 1, c = vy.size() - 1;
	a[++n] = {rr + 1, cc + 1};
	foa(i, 1, n) {
		int x = lower_bound(all(vx), a[i].fi) - vx.begin();
		int y = lower_bound(all(vy), a[i].se) - vy.begin();
		p[x][y] = 1;
		mx[x] = 1;
		if(x > 1) mx[x - 1] = vx[x] - vx[x - 2] - 1;
		my[y] = 1;
		if(y > 1) my[y - 1] = vy[y] - vy[y - 2] - 1;
	}
	// dbg(p, r, c, vx, mx, my)
	vector<int> res;
	foa(x, 1, r - 1) {
		foa(y, 1, c - 1) {
			if(p[x][y]) continue;
			int num = 0;
			queue<pii> q;
			q.push({x, y});
			p[x][y] = 1;
			while(q.size()) {
				auto t = q.front();
				q.pop();
				num += mx[t.fi] * my[t.se];
				// dbg(t, mx[t.fi], my[t.se])
				foa(i, 0, 3) {
					int nx = t.fi + dir[i][0], ny = t.se + dir[i][1];
					if(nx < 1 || nx >= r || ny < 1 || ny >= c || p[nx][ny]) continue;
					q.push({nx, ny});
					p[nx][ny] = 1;
				}
			}
			res.pb(num);
		}
	}
	sort(all(res));
	cout << res.size() << endl;
	foa(i, 0, res.size() - 1) cout << res[i] << " \n"[i == res.size() - 1];
	// cout << endl;
}
signed main()
{

#ifdef LOCAL
	freopen("read.in", "r", stdin);
	freopen("write.out", "w", stdout);
#endif

	std::ios::sync_with_stdio(false); std::cin.tie(0); std::cout.tie(0);

	init();

    int t = 0, num2 = 0;

    // t = 1;

    if (!t) cin >> t;
    while (t--) {
        cout<<"Case #"<<++num2<<":\n";
		solve();
    }
    return 0;

	int num1 = 0;
	//while (scanf("%d", &n) !=-1 && n)
    while (cin >> n && n) {
        //cout<<"Case "<<++num1<<": ";
		solve();
    }
    return 0;

}