Covering the stains
Problem Statement :
There is a huge blanket on your bed but unfortunately it has N stains. You cover them using a single, rectangular silk cloth. The silk is expensive, which is why the rectangular piece needs to have the least area as possible. You love this blanket and decide to minimize the area covering the stains. You buy some cleaning liquid to remove the stains but sadly it isn't enough to clean all of them. You can just remove exactly K stains. The rest of the stains need to be covered using a single, rectangular fragment of silk cloth. Let X denote the area of the smallest possible silk cloth that may cover all the stains originally. You need to find the number of different ways in which you may remove K stains so that the remaining N-K stains can be covered with silk of area strictly less than X (We are looking for any configuration that will reduce the cost). Assume that each stain is a point and that the rectangle is aligned parallel to the axes. Input Format The first line contains two integers N (1<=N<=1000) and K (0<=K<=N). Next follow N lines, one for each stain. Each line contains two integers in the form 'X Y', (0<=X,Y<100000), the coordinates of each stain into the blanket. Each pair of coordinates is unique. Output Format Output a single integer. The remainder of the division by 1000000007 of the answer.
Solution :
Solution in C :
In C++ :
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int MOD = 1000000007, MAXN = 1005;
int binom[MAXN][MAXN];
int N, K;
int X[MAXN], Y[MAXN];
int Xmin, Xmax, Ymin, Ymax;
pair<int, int> dude[4];
inline int calc(vector<pair<int, int> > vec) {
int num = 0;
for(int i = 0 ; i < N ; i++) {
for(int j = 0 ; j < (int)vec.size() ; j++) {
if (vec[j].first == 0 && X[i] == vec[j].second) {
num++;
break;
} else if (vec[j].first == 1 && Y[i] == vec[j].second) {
num++;
break;
}
}
}
if (num > K) {
return 0;
} else {
return binom[N - num][K - num];
}
}
int main() {
scanf("%d %d", &N, &K);
binom[0][0] = 1;
for(int i = 1 ; i <= N ; i++) {
binom[i][0] = 1;
for(int j = 1 ; j <= i ; j++) {
binom[i][j] = (binom[i - 1][j] + binom[i - 1][j - 1]) % MOD;
}
}
for(int i = 0 ; i < N ; i++) {
scanf("%d %d", &X[i], &Y[i]);
}
Xmin = X[0];
Xmax = X[0];
Ymin = Y[0];
Ymax = Y[1];
for(int i = 1 ; i < N ; i++) {
Xmin = min(Xmin, X[i]);
Xmax = max(Xmax, X[i]);
Ymin = min(Ymin, Y[i]);
Ymax = max(Ymax, Y[i]);
}
dude[0] = make_pair(0, Xmin);
dude[1] = make_pair(0, Xmax);
dude[2] = make_pair(1, Ymin);
dude[3] = make_pair(1, Ymax);
int ans = 0;
for(int mask = 1 ; mask < 16 ; mask++) {
int sgn = __builtin_popcount(mask);
vector<pair<int,int> > v;
for(int i = 0 ; i < 4 ; i++) {
if ((mask >> i) & 1) {
v.push_back(dude[i]);
}
}
if (sgn & 1) {
ans = (ans + calc(v)) % MOD;
} else {
ans = (ans - calc(v)) % MOD;
}
}
ans = (ans + MOD) % MOD;
ans = (ans + MOD) % MOD;
printf("%d\n", ans);
return 0;
}
In Java :
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
public class Solution {
private static Reader in;
private static PrintWriter out;
public static long mod = 1000000007;
public static long mod_exp(long b, long e) {
long r = 1;
while (e > 0) {
if ((e & 1) == 1) r = (r * b) % mod;
b = (b * b) % mod;
e >>= 1;
}
return r;
}
public static long[] fact, invfact;
public static void main(String[] args) throws IOException {
in = new Reader();
out = new PrintWriter(System.out, true);
fact = new long[1001];
invfact = new long[1001];
fact[0] = 1;
invfact[0] = 1;
for (int i = 1; i <= 1000; i++) {
fact[i] = (i * fact[i - 1]) % mod;
invfact[i] = mod_exp(fact[i], mod - 2);
}
int sx = 1 << 29, sy = 1 << 29, ex = -1, ey = -1;
int N = in.nextInt(), K = in.nextInt();
int[][] p = new int[N][2];
for (int i = 0; i < N; i++) {
p[i] = new int[2];
p[i][0] = in.nextInt();
p[i][1] = in.nextInt();
if (p[i][0] < sx) sx = p[i][0];
if (p[i][0] > ex) ex = p[i][0];
if (p[i][1] < sy) sy = p[i][1];
if (p[i][1] > ey) ey = p[i][1];
}
int[] b = new int[] {sx, ex, sy, ey};
int[] id = new int[] {0, 0, 1, 1};
long total = 0;
for (int mask = 1; mask < 1 << 4; mask++) {
int count = 0;
for (int i = 0; i < N; i++) {
boolean ok = false;
for (int j = 0; j < 4; j++) {
if (set(mask, j) && p[i][id[j]] == b[j])
ok = true;
}
if (ok) count++;
}
long sgn = Integer.bitCount(mask) % 2 == 1 ? 1 : (mod - 1);
total = (total + sgn * binom(N - count, K - count)) % mod;
}
out.println(total);
out.close();
System.exit(0);
}
public static boolean set(int m, int i) {
return ((m >> i) & 1) == 1;
}
public static long binom(int N, int K) {
if (K < 0 || K > N) return 0;
return fact[N] * invfact[K] % mod * invfact[N - K] % mod;
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[1024];
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n')
break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.')
while ((c = read()) >= '0' && c <= '9')
ret += (c - '0') / (div *= 10);
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}
In C :
#include <stdio.h>
#include <stdlib.h>
#define MOD 1000000007
long long modPow(long long a,int x);
long long modInverse(long long a);
long long C(int n,int k);
long long fac[1001];
long long ifac[1001];
int main(){
int N,K,maxx=-1,maxy=-1,minx=-1,miny=-1,u=0;
int d=0,l=0,r=0,ul=0,ur=0,dl=0,dr=0,i,t;
int *x,*y;
long long ans=0;
fac[0]=fac[1]=ifac[0]=ifac[1]=1;
for(i=2;i<1001;i++){
fac[i]=i*fac[i-1]%MOD;
ifac[i]=modInverse(fac[i]);
}
scanf("%d%d",&N,&K);
x=(int*)malloc(N*sizeof(int));
y=(int*)malloc(N*sizeof(int));
for(i=0;i<N;i++){
scanf("%d%d",x+i,y+i);
if(minx==-1 || x[i]<minx)
minx=x[i];
if(miny==-1 || y[i]<miny)
miny=y[i];
if(maxx==-1 || x[i]>maxx)
maxx=x[i];
if(maxy==-1 || y[i]>maxy)
maxy=y[i];
}
for(i=0;i<N;i++){
if(x[i]==minx)
l++;
if(x[i]==maxx)
r++;
if(y[i]==miny)
d++;
if(y[i]==maxy)
u++;
if(x[i]==minx && y[i]==miny)
dl++;
if(x[i]==minx && y[i]==maxy)
ul++;
if(x[i]==maxx && y[i]==miny)
dr++;
if(x[i]==maxx && y[i]==maxy)
ur++;
}
if(N==1){
if(K==0)
printf("0");
else
printf("1");
}
else if(minx==maxx || miny==maxy){
if(K>=1)
ans=(ans+C(N-1,K-1)*2%MOD)%MOD;
if(K>=2)
ans=(ans-C(N-2,K-2)+MOD)%MOD;
printf("%lld",ans);
}
else{
if(K>=u)
ans=(ans+C(N-u,K-u))%MOD;
if(K>=d)
ans=(ans+C(N-d,K-d))%MOD;
if(K>=l)
ans=(ans+C(N-l,K-l))%MOD;
if(K>=r)
ans=(ans+C(N-r,K-r))%MOD;
t=u+d;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=u+l-ul;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=u+r-ur;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=d+l-dl;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=d+r-dr;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=l+r;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
t=u+d+l-ul-dl;
if(K>=t)
ans=(ans+C(N-t,K-t))%MOD;
t=u+d+r-ur-dr;
if(K>=t)
ans=(ans+C(N-t,K-t))%MOD;
t=u+r+l-ul-ur;
if(K>=t)
ans=(ans+C(N-t,K-t))%MOD;
t=r+d+l-dl-dr;
if(K>=t)
ans=(ans+C(N-t,K-t))%MOD;
t=u+d+l+r-ul-ur-dl-dr;
if(K>=t)
ans=(ans-C(N-t,K-t)+MOD)%MOD;
printf("%lld",ans);
}
return 0;
}
long long modPow(long long a,int x){
long long res = 1;
while(x>0){
if(x%2)
res=res*a%MOD;
a=a*a%MOD;
x>>=1;
}
return res;
}
long long modInverse(long long a){
return modPow(a,MOD-2);
}
long long C(int n,int k){
return fac[n]*ifac[k]%MOD*ifac[n-k]%MOD;
}
In Python3 :
def stain_combos(n, k):
p = 1000000007
if k < 0:
return 0
out = 1
for i in range(n-k+1, n+1):
out = (out * i) % p
denom = 1
for i in range(1, k+1):
denom = (denom * i) % p
denom = pow(denom, p-2, p)
return (out*denom)%p
def solve(n, k, stains):
print(size_decrease(stains, k))
from itertools import combinations
from functools import reduce
def size_decrease(stains, k):
min_x = min([p.x for p in stains])
max_x = max([p.x for p in stains])
min_y = min([p.y for p in stains])
max_y = max([p.y for p in stains])
top = {p for p in stains if p.x==min_x}
bot = {p for p in stains if p.x==max_x}
left = {p for p in stains if p.y==min_y}
right = {p for p in stains if p.y==max_y}
out = 0
for i in range(1, 5):
for sides in combinations([top, bot, left, right], i):
removed = reduce(lambda x,y: x.union(y), sides)
length = len(removed)
out += ((-1)**(i+1)) * stain_combos(len(stains)-length, k-length)
return out%1000000007
from collections import namedtuple
point = namedtuple('point', ['x','y'])
n, k = [int(x) for x in input().split(" ")]
stains = []
for _ in range(n):
stains.append(point(*[int(number) for number in input().split(" ")]))
solve(n, k, stains)
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