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

Keywords: Java network

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

The meaning of this question is that in a graph of nn square, there are m rectangular squares which are black and the others are white. It takes min(a,b) to turn an ab rectangle white. It can be seen from this that it is the same price to make the whole row or column white and not crazy white at a time.
This is a problem of minimum point coverage. Minimum Point Coverage = Maximum Network Flow in Bipartite Graph. Points are rows and columns. The edge is the black block in the picture. However, because the number of n may be very large, leading to a large graph, it is certainly not possible to directly use a bipartite graph. It needs to be discretized. Merge the same rows and columns to form a new capacity.
After discretization, the source and sink points are established and connected with rows and columns respectively. Capacity is the capacity formed after merging. The side capacity composed of rows and columns is infinite. Run the maximum flow algorithm to get the result after the graph is built.

The first network stream we did had to be remembered...

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

public class Main {
static StreamTokenizer st = new StreamTokenizer(new BufferedInputStream(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;
static int et = 1, h[] = new int, d[] = new int, ss, tt;
static int a[] = new int, b[] = new int, c[][] = new int;
static int x1[] = new int, y1[] = new int, x2[] = new int, y2[] = new int;

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] = x1[i];
a[++a] = x2[i] + 1;
b[++b] = y1[i];
b[++b] = y2[i] + 1;
}
Arrays.sort(a, 1, a + 1);
a = unique(a, 1, a + 1);

Arrays.sort(b, 1, b + 1);
b = unique(b, 1, b + 1) ;
for (int i = 1; i <= m; i++) {
x1[i] = lower_bound(a, 1, a + 1, x1[i]);
x2[i] = lower_bound(a, 1, a + 1, x2[i] + 1) - 1;
y1[i] = lower_bound(b, 1, b + 1, y1[i]);
y2[i] = lower_bound(b, 1, b + 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 + b + 1;
for (int i = 1; i <= a; i++)
for (int j = 1; j <= b; j++)
if (c[i][j] != 0)
jb(i, a + j, (int) 2e9);
for (int i = 1; i < a; i++)
jb(ss, i, a[i + 1] - a[i]);
for (int i = 1; i < b; i++)
jb(a + 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>();
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;
}
}
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;
}
}

Posted by yurko on Wed, 09 Oct 2019 01:06:01 -0700