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Ants and Anteaters CodeChef Solution
Problem -Ants and Anteaters CodeChef Solution
“This website is dedicated for CodeChef solution where we will publish right solution of all your favourite CodeChef problems along with detailed explanatory of different competitive programming concepts and languages.
Ants and Anteaters CodeChef Solution in C++14
“#include <bits/stdc++.h>
using namespace std;
//order_of_key (k) : Number of items strictly smaller than k .
//find_by_order(k) : K-th element in a set (counting from zero).
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
typedef tree<int,null_type,less<int>,rb_tree_tag,
tree_order_statistics_node_update> indexed_set;
const double pi = 3.14159265358979;
const long long mod=1e9+7;
#define inf 0x3f3f3f3f
#define endl "\n";
#define int long long
vector<int> sieve(int n) {int*arr = new int[n + 1](); vector<int> vect; for (int i = 2; i <= n; i++)if (arr[i] == 0) {vect.push_back(i); for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return vect;}
int binpow(int a,int b,int m){a=a%m; int res=1;while(b>0){ if(b&1) res=res*a%m; a=a*a%m; b>>=1; }return res;}
bool isvowel(char v) { return (0x208222>>(v&0x1f))&1;}
bool ispowerof2(int n){ if(n<=0) return false; if((n&(n-1))==0) return true; return false;}
bool onesInBinaryRep(int n){ int cnt=0; while(n>0){ n=n&(n-1); cnt++;} if(cnt%2==0) return true; return false; }
int ceil(int x,int y){ return (x+y-1)/y;}
void solve(){
int n,m; cin>>n>>m;
vector<vector<char>> v(n,vector<char>(m));
for(int i=0; i<n; i++){
string s; cin>>s;
for(int j=0; j<m; j++)
v[i][j]=s[j];
}
int ans=0;
queue<pair<pair<int,int>,char>> q;
for(int i=0; i<n; i++){
for(int j=0; j<m; j++){
if(v[i][j]!='-' && v[i][j]!='#'){
q.push({{i,j},v[i][j]});
}
}
}
while(!q.empty()){
int sz=q.size();
map<pair<int,int>,int> mp;
for(int i=0; i<sz; i++){
pair<pair<int,int>,char> p=q.front(); q.pop();
int x=p.first.first; int y=p.first.second;
char z=p.second;
if(z=='D'){
if(x+1<n){
if(v[x+1][y]!='#'){
mp[{x+1,y}]++;
q.push({{x+1,y},z});
}
}
}
else if(z=='U'){
if(x-1>=0){
if(v[x-1][y]!='#'){
mp[{x-1,y}]++;
q.push({{x-1,y},z});
}
}
}
else if(z=='L'){
if(y-1>=0){
if(v[x][y-1]!='#'){
mp[{x,y-1}]++;
q.push({{x,y-1},z});
}
}
}
else if(z=='R'){
if(y+1<m){
if(v[x][y+1]!='#'){
mp[{x,y+1}]++;
q.push({{x,y+1},z});
}
}
}
}
for(auto it : mp){
if(it.second>=2){
ans+=it.second*(it.second-1);
}
}
}
cout<<ans/2<<endl;
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t; cin>>t;
while(t--){
solve();
}
}Ants and Anteaters CodeChef Solution in PYTH 3
“# cook your dish here
test = int(input())
for _ in range(test):
R, C = list(map(int, input().split()))
grid = []
R_list = []
L_list = []
U_list = []
D_list = []
pairs = 0
for i in range(R):
row = list(input())
R_list.append(list((i for i,value in enumerate(row) if value == "R")))
L_list.append(list((i for i,value in enumerate(row) if value == "L")))
U_list.append(list((i for i,value in enumerate(row) if value == "U")))
D_list.append(list((i for i,value in enumerate(row) if value == "D")))
grid.append(row)
for i in range(max(R, C)):
for j in range(len(R_list)):
for x in range(len(R_list[j])):
if R_list[j][x] + 1 >= C:
R_list[j][x] = "-"
elif grid[j][R_list[j][x] + 1] == "#":
R_list[j][x] = "-"
else:
R_list[j][x] = R_list[j][x] + 1
R_list[j] = list(filter(("-").__ne__, R_list[j]))
for j in range(len(L_list)):
for x in range(len(L_list[j])):
if L_list[j][x] - 1 < 0:
L_list[j][x] = "-"
elif grid[j][L_list[j][x] - 1] == "#":
L_list[j][x] = "-"
else:
L_list[j][x] = L_list[j][x] - 1
L_list[j] = list(filter(("-").__ne__, L_list[j]))
for j in range(len(U_list)):
for x in range(len(U_list[j])):
if j == 0:
U_list[j][x] = "-"
elif grid[j - 1][U_list[j][x]] == "#":
U_list[j][x] = "-"
else:
U_list[j - 1].append(U_list[j][x])
U_list[j][x] = "-"
U_list[j] = list(filter(("-").__ne__, U_list[j]))
for j in range(len(D_list)):
for x in range(len(D_list[j])):
if isinstance(D_list[j][x], str):
continue
if j == R - 1:
D_list[j][x] = "-"
elif grid[j + 1][D_list[j][x]] == "#":
D_list[j][x] = "-"
else:
D_list[j + 1].append(str(D_list[j][x]))
D_list[j][x] = "-"
D_list[j] = list(filter(("-").__ne__, D_list[j]))
for k in range(R):
D_list[k] = list(map(int, D_list[k]))
for k in range(R):
if len(R_list[k]) != 0 and len(L_list[k]) != 0:
s1 = set(R_list[k])
s2 = set(L_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
if len(R_list[k]) != 0 and len(D_list[k]) != 0:
s1 = set(R_list[k])
s2 = set(D_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
if len(R_list[k]) != 0 and len(U_list[k]) != 0:
s1 = set(R_list[k])
s2 = set(U_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
if len(L_list[k]) != 0 and len(D_list[k]) != 0:
s1 = set(L_list[k])
s2 = set(D_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
if len(L_list[k]) != 0 and len(U_list[k]) != 0:
s1 = set(L_list[k])
s2 = set(U_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
if len(U_list[k]) != 0 and len(D_list[k]) != 0:
s1 = set(U_list[k])
s2 = set(D_list[k])
set1 = s1.intersection(s2)
pairs += len(list(set1))
print(pairs)
Ants and Anteaters CodeChef Solution in C
“#include <stdio.h>
char s[50][50];
int r,c;
int valid(int i,int j)
{
if(i>=0&&i<=r-1&&j>=0&&j<=c-1&&s[i][j]!='#')
return 1;
else
return 0;
}
int main()
{
int t,i,j,k,up,down,right,left,l,meet;
scanf("%d",&t);
for(k=0;k<t;k++)
{
scanf("%d%d",&r,&c);
for(j=0;j<r;j++)
scanf("%s",s[j]);
meet=0;
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
{
up=1;
down=1;
right=1;
left=1;
l=1;
if(s[i][j]=='#')
continue;
while(up||down||right||left)
{
if(!valid(i-l,j))
up=0;
if(!valid(i+l,j))
down=0;
if(!valid(i,j-l))
left=0;
if(!valid(i,j+l))
right=0;
if(up&&down&&s[i-l][j]=='D'&&s[i+l][j]=='U')
meet++;
if(up&&right&&s[i-l][j]=='D'&&s[i][j+l]=='L')
meet++;
if(up&&left&&s[i-l][j]=='D'&&s[i][j-l]=='R')
meet++;
if(down&&left&&s[i+l][j]=='U'&&s[i][j-l]=='R')
meet++;
if(down&&right&&s[i+l][j]=='U'&&s[i][j+l]=='L')
meet++;
if(right&&left&&s[i][j+l]=='L'&&s[i][j-l]=='R')
meet++;
l++;
}
}
}
printf("%d\n",meet);
}
return 0;
}Ants and Anteaters CodeChef Solution in JAVA
“import java.io.*;
import java.util.*;
class anteater {
public static void main(String[] args)throws IOException {
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
for(int z=1;z<=t;z++){
int n=sc.nextInt();
int m=sc.nextInt();
char ch[][]=new char[n][m];
int count[][][]=new int[n][m][51];
for(int i=0;i<n;i++){
String s=sc.next();
for(int j=0;j<m;j++){
ch[i][j]=s.charAt(j);
}
}
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(ch[i][j]=='R'){
update(count,0,1,n,m,i,j,ch);
}
else if(ch[i][j]=='L'){
update(count,0,-1,n,m,i,j,ch);
}
else if(ch[i][j]=='U'){
update(count,-1,0,n,m,i,j,ch);
}
else if(ch[i][j]=='D') {
update(count,1,0,n,m,i,j,ch);
}
}
}
int pairs=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(ch[i][j]!='#')
{
//System.out.println(i+" "+j);
for(int k=0;k<=50;k++){
//System.out.print(count[i][j][k]+" ");
int tt=count[i][j][k];
pairs+=((tt*(tt-1))/2);
}
//System.out.println();
}
}
}
System.out.println(pairs);
}
}
static void update(int a[][][],int r,int c,int n,int m,int i,int j,char ch[][]){
int cc=0;
while(i<n && j<m && i>=0 && j>=0){
if(ch[i][j]=='#'){
break;
}
a[i][j][cc]++;
i+=r;
j+=c;
cc++;
}
}
}Ants and Anteaters CodeChef Solution in PYPY 3
“from collections import Counter
def binomialCoef(n, k):
C = [[0 for x in range(k+1)] for x in range(n+1)]
# Calculate value of Binomial Coefficient in bottom up manner
for i in range(n+1):
for j in range(min(i, k)+1):
# Base Cases
if j == 0 or j == i:
C[i][j] = 1
# Calculate value using previously stored values
else:
C[i][j] = C[i-1][j-1] + C[i-1][j]
return C[n][k]
for _ in range(int(input())):
r,c=map(int,input().split())
d=dict()
grid=[None]*r
for i in range(r):
grid[i]=input()
for i in range(r):
for j in range(c):
if grid[i][j]=="U":
kk=i
sec=0
while kk>=0:
if grid[kk][j]=="#":
break
try:
d["r"+str(kk)+"c"+str(j)].append(sec)
except:
d["r"+str(kk)+"c"+str(j)]=[sec]
sec+=1
kk-=1
if grid[i][j]=="D":
kk=i
sec=0
while kk<r:
if grid[kk][j]=="#":
break
try:
d["r"+str(kk)+"c"+str(j)].append(sec)
except:
d["r"+str(kk)+"c"+str(j)]=[sec]
sec+=1
kk+=1
if grid[i][j]=="R":
kk=j
sec=0
while kk<c:
if grid[i][kk]=="#":
break
try:
d["r"+str(i)+"c"+str(kk)].append(sec)
except:
d["r"+str(i)+"c"+str(kk)]=[sec]
sec+=1
kk+=1
if grid[i][j]=="L":
kk=j
sec=0
while kk>=0:
if grid[i][kk]=="#":
break
try:
d["r"+str(i)+"c"+str(kk)].append(sec)
except:
d["r"+str(i)+"c"+str(kk)]=[sec]
sec+=1
kk-=1
ans=0
for i in d.values():
f=dict(Counter(i))
for j in f.values():
ans+=binomialCoef(j,2)
print(ans)
Ants and Anteaters CodeChef Solution in PYTH
“from collections import defaultdict
t = int(raw_input())
for i in range(t):
st = raw_input().split()
R = int(st[0])
C = int(st[1])
A = []
S = set()
for y in range(R+2):
S.add((y,0))
S.add((y,C+1))
# endfor y
for x in range(C+2):
S.add((0,x))
S.add((0,R+1))
# endfor x
for y in range(1,R+1):
st = raw_input().strip()
for x in range(1,C+1):
ch = st[x-1]
if ch == '#':
S.add((y,x))
elif ch == 'U':
A.append([y,x,-1,0])
elif ch == 'D':
A.append([y,x,1,0])
elif ch == 'L':
A.append([y,x,0,-1])
elif ch == 'R':
A.append([y,x,0,1])
# endif
# endfor x
# endfor y
num = min(R,C) -1
num = max(num,R/2,C/2)
tot = 0
sz = len(A)
for k in range(num):
G= [[0 for x in range(C+1)] for y in range(R+1)]
D = defaultdict(int)
p = 0
while p < sz:
y = A[p][0]+A[p][2]
x = A[p][1]+A[p][3]
if not (y,x) in S:
tot += D[y,x]
D[y,x] += 1
A[p][0] = y
A[p][1] = x
else:
del(A[p])
sz -= 1
p -= 1
# endif
p += 1
# endwhile
# endfor k
print tot
# endfor i
Ants and Anteaters CodeChef Solution in GO
“package main
import "fmt"
import "io"
import "math"
import "os"
func main() {
T := ScanInt(1, 10)
for t := 1; t <= T; t++ {
solve()
}
}
var g1, g2 [50][50]byte
const (
up = 1 << iota
down
right
left
eater
)
var bits = []int{
0, 1, 1, 2,
1, 2, 2, 3,
1, 2, 2, 3,
2, 3, 3, 4,
}
func solve() {
NewLine()
R := ScanInt(1, 50)
C := ScanInt(1, 50)
grid := &g1
for r := 0; r < R; r++ {
NewLine()
row := ScanString(C, C)
for c := 0; c < C; c++ {
switch row[c] {
case 'U':
grid[r][c] = up
case 'D':
grid[r][c] = down
case 'R':
grid[r][c] = right
case 'L':
grid[r][c] = left
case '#':
grid[r][c] = eater
case '-':
grid[r][c] = 0
default:
panic("bad input")
}
}
}
next := &g2
count := 0
for moving := true; moving; {
moving = false
for r := 0; r < R; r++ {
for c := 0; c < C; c++ {
if grid[r][c] == eater {
next[r][c] = eater
} else {
next[r][c] = 0
}
}
}
enter := func(r, c int, dir byte) {
if 0 <= r && r < R && 0 <= c && c < C && next[r][c] != eater {
next[r][c] |= dir
if next[r][c] != 0 {
moving = true
}
}
}
for r := 0; r < R; r++ {
for c, cell := range grid[r][:C] {
if cell & up != 0 {
enter(r - 1, c, up)
}
if cell & down != 0 {
enter(r + 1, c, down)
}
if cell & left != 0 {
enter(r, c - 1, left)
}
if cell & right != 0 {
enter(r, c + 1, right)
}
}
}
grid, next = next, grid
for r := 0; r < R; r++ {
for _, cell := range grid[r][:C] {
if cell < eater {
n := bits[cell]
count += n * (n - 1) / 2
}
}
}
}
fmt.Println(count)
}
func ScanInt(low, high int) int {
return int(ScanInt64(int64(low), int64(high)))
}
func NewLine() {
if CheckInput {
for n, b := range RemainingInput {
if b != ' ' && b != '\t' && b != '\r' {
Assert(b == '\n')
RemainingInput = RemainingInput[n+1:]
return
}
}
Assert(false)
}
}
func ScanString(short, long int) string {
return string(ScanBytes(short, long))
}
func ScanInt64(low, high int64) int64 {
x := Btoi(ScanToken())
Assert(low <= x && x <= high || !CheckInput)
return x
}
func Assert(condition bool) {
if !condition {
panic("assertion failed")
}
}
func ScanBytes(short, long int) []byte {
token := ScanToken()
Assert(short <= len(token) && len(token) <= long)
return token
}
func Btoi(s []byte) int64 {
if s[0] == '-' {
x := Btou(s[1:])
Assert(x <= - math.MinInt64)
return - int64(x)
} else {
x := Btou(s)
Assert(x <= math.MaxInt64)
return int64(x)
}
}
var RemainingInput []byte
var CheckInput = true
func ScanToken() []byte {
chunk := 1 << 16
skipws: for {
for n, b := range RemainingInput {
if b != ' ' && b != '\t' && b != '\r' {
if b == '\n' && CheckInput {
panic("ScanToken: reached end of line")
}
RemainingInput = RemainingInput[n:]
break skipws
}
}
if len(RemainingInput) == 0 {
RemainingInput = make([]byte, chunk)
n, e := os.Stdin.Read(RemainingInput)
if e != nil && e != io.EOF {
panic(e)
}
if n == 0 {
panic("ScanToken: reached EOF")
}
RemainingInput = RemainingInput[:n]
}
}
n := 1
for {
if n == len(RemainingInput) {
ri := make([]byte, 2 * n + chunk)
copy(ri, RemainingInput)
k, e := os.Stdin.Read(ri[n:])
if e != nil && e != io.EOF {
panic(e)
}
RemainingInput = ri[:n+k]
if k == 0 {
if CheckInput {
panic("ScanToken: EOF in middle of token")
} else {
break // and pray
}
}
}
b := RemainingInput[n]
if b == ' ' || b == '\t' || b == '\r' || b == '\n' {
break
}
n++
}
token := RemainingInput[:n]
RemainingInput = RemainingInput[n:]
return token
}
func Btou(s []byte) uint64 {
Assert(len(s) > 0)
var x uint64
for _, d := range s {
Assert('0' <= d && d <= '9')
d -= '0'
if x >= math.MaxUint64 / 10 {
Assert(x == math.MaxUint64 / 10 && d <= math.MaxUint64 % 10)
}
x = x * 10 + uint64(d)
}
return x
}Ants and Anteaters CodeChef Solution Review:
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