Codeforces Round #576 (Div. 1) E. Rectangle Painting 2

Rectangle Painting 2

There is a square grid of size n×n. Some cells are colored in black, all others are colored in white. In one operation you can select some rectangle and color all its cells in white. It costs min(h,w) to color a rectangle of size h×w. You are to make all cells white for minimum total cost.

The square is large, so we give it to you in a compressed way. The set of black cells is the union of m rectangles.

Input

The first line contains two integers n and m (1≤n≤109, 0≤m≤50) — the size of the square grid and the number of black rectangles.

Each of the next m lines contains 4 integers xi1 yi1 xi2 yi2 (1≤xi1≤xi2≤n, 1≤yi1≤yi2≤n) — the coordinates of the bottom-left and the top-right corner cells of the i-th black rectangle.

The rectangles may intersect.

Output

Print a single integer — the minimum total cost of painting the whole square in white.

Examples

input

10 2
4 1 5 10
1 4 10 5

output

4

input

7 6
2 1 2 1
4 2 4 3
2 5 2 5
2 3 5 3
1 2 1 2
3 2 5 3

output

3

該題的意思是，在一個nn正方形的圖中，有m個矩形的方塊是黑色的，其它的是白色的。而使一個ab的矩形變成白色需要min(a,b)的代價。從這裏可以看出可以每次使整行或者整列變成白色和不瘋白色是一樣的代價。
這樣就是一個最小點覆蓋的問題。最小點覆蓋=二分圖中的最大網絡流。點是行和列。邊是圖中的黑色色塊。但由於n的數可能很大，導致圖很大，直接用二分圖肯定不行的。需要進行離散化。將一樣的行和列進行合併，組成新的容量。
離散之後建立源點和匯點，分別和行列相連，容量就是合併之後形成的容量，行和列相連組成的邊容量無窮大。圖建好之後跑最大流算法得出結果。

做的第一個網絡流怎麼也得留個紀念。。。。
java AC代碼

import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main {
	static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(System.in));
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter pr = new PrintWriter(new BufferedOutputStream(System.out));
	static Scanner sc = new Scanner(System.in);
//		long tic = System.currentTimeMilis();
//		long toc = System.currentTimeMillis();
//		System.out.println("Elapsed time: " + (toc - tic) + " ms");
	static Node e[] = new Node[10000001];
	static int et = 1, h[] = new int[10001], d[] = new int[10001], ss, tt;
	static int a[] = new int[101], b[] = new int[101], c[][] = new int[101][101];
	static int x1[] = new int[101], y1[] = new int[101], x2[] = new int[101], y2[] = new int[101];

	public static void main(String[] args) {
		int n = nextInt(), m = nextInt();
		for (int i = 1; i <= m; i++) {
			x1[i] = nextInt();
			y1[i] = nextInt();
			x2[i] = nextInt();
			y2[i] = nextInt();
			a[++a[0]] = x1[i];
			a[++a[0]] = x2[i] + 1;
			b[++b[0]] = y1[i];
			b[++b[0]] = y2[i] + 1;
		}
		Arrays.sort(a, 1, a[0] + 1);
		a[0] = unique(a, 1, a[0] + 1);
 
		Arrays.sort(b, 1, b[0] + 1);
		b[0] = unique(b, 1, b[0] + 1) ;
		for (int i = 1; i <= m; i++) {
			x1[i] = lower_bound(a, 1, a[0] + 1, x1[i]);
			x2[i] = lower_bound(a, 1, a[0] + 1, x2[i] + 1) - 1;
			y1[i] = lower_bound(b, 1, b[0] + 1, y1[i]);
			y2[i] = lower_bound(b, 1, b[0] + 1, y2[i] + 1) - 1;
			for (int u = x1[i]; u <= x2[i]; u++)
				for (int v = y1[i]; v <= y2[i]; v++)
					c[u][v] = 1;
		}
		ss = 0;
		tt = a[0] + b[0] + 1;
		for (int i = 1; i <= a[0]; i++)
			for (int j = 1; j <= b[0]; j++)
				if (c[i][j] != 0)
					jb(i, a[0] + j, (int) 2e9);
		for (int i = 1; i < a[0]; i++)
			jb(ss, i, a[i + 1] - a[i]);
		for (int i = 1; i < b[0]; i++)
			jb(a[0] + i, tt, b[i + 1] - b[i]);
		int ans = 0;
		while (bfs()) {
			ans += dfs(ss, (int) 2e9);
		}
		System.out.printf("%d\n", ans);
	}
 
	private static int lower_bound(int[] arr, int l, int r, int it) {
		int m = (l+r)>>1;
		while(l<=r) {
			if(it > arr[m]) {
				l = m+1;
				m = (l+r)>>1;
			}else if(it<arr[m]) {
				r = m-1;
				m = (l+r)>>1;
			}else {
				return m;
			}
		}
		return m;
	}
 
	private static int unique(int[] a2, int a, int b) {
		int te = a;
		for (int i = a + 1; i < b; i++) {
			if (a2[i] != a2[te]) {
				a2[++te] = a2[i];
			}
		}
		return te - a + 1;
	}
 
	static void jb(int u, int v, int f) {
		e[++et] = new Node(v, f, h[u]);
		h[u] = et;
		e[++et] = new Node(u, 0, h[v]);
		h[v] = et;
	}
 
	static boolean bfs() {
		for (int i = ss; i <= tt; i++) {
			d[i] = 0;
		}
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(ss);
		d[ss] = 1;
		while (!q.isEmpty()) {
			int u = q.poll();
			for (int i = h[u]; i != 0; i = e[i].nx)
				if (e[i].f != 0 && d[e[i].v] == 0) {
					d[e[i].v] = d[u] + 1;
					q.add(e[i].v);
				}
		}
		return d[tt] != 0;
	}
 
	static int dfs(int u, int f) {
		if (f == 0 || u == tt)
			return f;
		int F = 0;
		for (int i = h[u]; i != 0; i = e[i].nx)
			if (d[e[i].v] == d[u] + 1) {
				int ff = dfs(e[i].v, Math.min(e[i].f, f));
				e[i].f -= ff;
				e[i ^ 1].f += ff;
				f -= ff;
				F += ff;
				if (f == 0)
					break;
			}
		return F;
	}
 
	static long pow(long a, long b, long mod) {
		if (b == 0)
			return 1;
		if (b == 1)
			return a % mod;
		if ((b & 1) == 0)
			return pow(a * a % mod, b >> 1, mod) % mod;
		else
			return a * pow(a * a % mod, b >> 1, mod) % mod;
	}

	static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		} else {
			return gcd(b, a % b);
		}
	}

	static int nextInt() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (int) st.nval;
	}

	static double nextDouble() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.nval;
	}

	static String next() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return st.sval;
	}

	static long nextLong() {
		try {
			st.nextToken();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return (long) st.nval;
	}
}
class Node {
	int v, f, nx;

	public Node() {
	}

	public Node(int v, int f, int nx) {
		this.v = v;
		this.f = f;
		this.nx = nx;
	}
}