Physical Address

304 North Cardinal St.
Dorchester Center, MA 02124

# Maze of Digits CodeChef Solution

## Maze of Digits CodeChef Solution in C++17

``````#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <string>
using namespace std;

const int MAXN = 105;
string map[MAXN];
int dir[5] = {1, 0, -1, 0, 1};
int f[MAXN][MAXN][MAXN];
int T, N, M, X, sx, sy;

int bfs(int y, int x, int v) {
int nx, ny, nv;
queue<int> Q;
memset(f, 0x33, sizeof(f));
f[y][x][v] = 0;
Q.push(y), Q.push(x), Q.push(v);

while(!Q.empty()) {
y = Q.front(), Q.pop();
x = Q.front(), Q.pop();
v = Q.front(), Q.pop();

for(int i = 0; i < 4; i++) {
ny = y + dir[i];
nx = x + dir[i + 1];
if(ny >= 0 && ny <= N && nx >= 0 && nx <= M)
if(map[ny][nx] != '#') {
nv = v + map[ny][nx] - '0';
if(nv == X) return f[y][x][v] + 1;
if(nv > X) continue;
if(f[ny][nx][nv] > f[y][x][v] + 1) {
f[ny][nx][nv] = f[y][x][v] + 1;
Q.push(ny), Q.push(nx), Q.push(nv);
}
}
}
}
return -1;
}

int main() {
cin >> T;
while(T--) {
cin >> N >> M;
for(int i = 0; i <= N; i++) {
cin >> map[i];
for(int j = 0; j <= M; j++) {
if(map[i][j] == '*') sy = i, sx = j, map[i][j] = '0';
else if(map[i][j] == '.') map[i][j] = '0';
}
}
cin >> X;
cout << bfs(sy, sx, 0) << endl;
}
return 0;
}``````

## Maze of Digits CodeChef Solution in C++14

``````#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = 3.14159265358979323846;  /* pi */
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 3030303;
const ll mod = 1e9 + 7;

// typedef
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef deque<ll> dqll;
typedef deque<pll> dqpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) (a & (1ll<<i))
#define BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\n"
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"Possible\n";
#define pimpossible cout<<"Impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n){cout<<v[i];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
void readf() {freopen("", "rt", stdin);}
template<typename T>
void readv(vector<T>& v){rep(i,0,sz(v)) cin>>v[i];}
template<typename T, typename U>
void readp(pair<T,U>& A) {cin>>A.first>>A.second;}
template<typename T, typename U>
void readvp(vector<pair<T,U>>& A) {rep(i,0,sz(A)) readp(A[i]); }
void readvall3(vall3& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2];}
void readvall5(vall5& A) {rep(i,0,sz(A)) cin>>A[i][0]>>A[i][1]>>A[i][2]>>A[i][3]>>A[i][4];}
void readvvll(vvll& A) {rep(i,0,sz(A)) readv(A[i]);}

struct Combination {
vll fac, inv;
ll n, MOD;

ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
inv = fac = vll(n,1);
rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
}

ll fact(ll n) {return fac[n];}
ll nCr(ll n, ll r) {
if(n < r or n < 0 or r < 0) return 0;
return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
}
};

struct Matrix {
ll r,c;
vvll matrix;
Matrix(ll r, ll c, ll v = 0): r(r), c(c), matrix(vvll(r,vll(c,v))) {}

Matrix operator*(const Matrix& B) const {
Matrix res(r, B.c);
rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
res.matrix[i][j] = (res.matrix[i][j] + matrix[i][k] * B.matrix[k][j] % mod) % mod;
}
return res;
}

Matrix copy() {
Matrix copy(r,c);
copy.matrix = matrix;
return copy;
}

Matrix pow(ll n) {
assert(r == c);
Matrix res(r,r);
Matrix now = copy();
rep(i,0,r) res.matrix[i][i] = 1;
while(n) {
if(n & 1) res = res * now;
now = now * now;
n /= 2;
}
return res;
}
};

// geometry data structures
template <typename T>
struct Point {
T y,x;
Point(T y, T x) : y(y), x(x) {}
Point(pair<T,T> p) : y(p.first), x(p.second) {}
Point() {}
void input() {cin>>y>>x;}
friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }
bool operator<(const Point &other) const {if (x == other.x) return y < other.y;return x < other.x;}
Point<T> rotate(Point<T> center, ld angle) {
ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
ld y = this->y - center.y;
ld x = this->x - center.x;

return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
}
ld distance(Point<T> other) {
T dy = abs(this->y - other.y);
T dx = abs(this->x - other.x);
return sqrt(dy * dy + dx * dx);
}

T norm() { return x * x + y * y; }
};

template<typename T>
struct Line {
Point<T> A, B;
Line(Point<T> A, Point<T> B) : A(A), B(B) {}
Line() {}

void input() {
A = Point<T>();
B = Point<T>();
A.input();
B.input();
}

T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
T res = a.x * b.y + b.x * c.y + c.x * a.y;
res -= (a.x * c.y + b.x * a.y + c.x * b.y);
return res;
}

bool isIntersect(Line<T> o) {
T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
if (p1p2 == 0 && p3p4 == 0) {
pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
if (p1 > p2) swap(p2, p1);
if (p3 > p4) swap(p3, p4);
return p3 <= p2 && p1 <= p4;
}
return p1p2 <= 0 && p3p4 <= 0;
}

pair<bool,Point<ld>> intersection(Line<T> o) {
if(!this->intersection(o)) return {false, {}};
ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
}

//@formula for : y = ax + b
//@return {a,b};
pair<ld, ld> formula() {
T y1 = A.y, y2 = B.y;
T x1 = A.x, x2 = B.x;
if(y1 == y2) return {1e9, 0};
if(x1 == x2) return {0, 1e9};
ld a = 1. * (y2 - y1) / (x2 - x1);
ld b = -x1 * a + y1;
return {a, b};
}
};

template<typename T>
struct Circle {
Point<T> center;
T radius;
Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
Circle(Point<T> center, T radius) : center(center), radius(radius) {}
Circle() {}

void input() {
center = Point<T>();
center.input();
cin>>radius;
}

bool circumference(Point<T> p) {
return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
}

bool intersect(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
}

bool include(Circle<T> c) {
T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
return d <= radius * radius;
}
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) { n%=MOD; if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

ll vis[101][101][101];

ll bfs(vs& A, ll sy, ll sx, ll k) {
if(k == 1) return 0;
queue<all3> q;
q.push({0,sy,sx});
vis[0][sy][sx] = 0;
ll dy[4]{-1,0,1,0},dx[4]{0,1,0,-1};
ll n = sz(A), m = sz(A[0]);
while(sz(q)) {
auto [val, y, x] = q.front(); q.pop();
rep(i,0,4) {
ll ny = y + dy[i], nx = x + dx[i];
if(0 <= ny and ny < n and 0 <= nx and nx < m and A[ny][nx] != '#') {
ll cost = 1 + vis[val][y][x];
ll v = val + (isdigit(A[ny][nx]) ? A[ny][nx] - '0' : 0);
if(v > k) continue;
if(vis[v][ny][nx] <= cost) continue;
vis[v][ny][nx] = cost;
if(v == k)
return cost;
q.push({v,ny,nx});
}
}
}
return -1;
}

ll solve(vs A, ll x) {
INF(vis);
rep(i,0,sz(A)) rep(j,0,sz(A[i])) {
if(A[i][j] == '*') return bfs(A,i,j,x);
}
return -1;
}
int main() {
Code By Sumfi
cout.precision(12);
ll tc = 1;
cin>>tc;
rep(i,1,tc+1) {
ll n,m,x;
cin>>n>>m;
vs A(n + 1);
readv(A);
cin>>x;
print(solve(A,x));
}
return 0;
}``````

## Maze of Digits CodeChef Solution in C

``````#include <stdio.h>
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MX 102
#define true 1
#define false 0
#define MXQ 10005

const int INF = 100000005;
int R,C,D,dx[]={0,0,1,-1},dy[]={1,-1,0,0};
int G[MX][MX];
int vis[MX][MX];
int DigInd[MX][MX];
int DigsX[MX], DigsY[MX], DigsN[MX];
int Dis[MX][MX];
int dp[MX][MX];
int Q[MXQ];

void bfsStart(int sx,int sy)
{
DigInd[sx][sy]=0;
DigsX[0]=sx; DigsY[0]=sy; DigsN[0]=0;
int i,j,k,cx,cy,cs,nx,ny,ns,x;

for(i=0;i<=R;i++) for(j=0;j<=C;j++) vis[i][j]=false;

D=1;
int H=0,T=0;
Q[T++]=sx+1000*sy;
vis[sx][sy]=true;

while(H!=T)
{
x=Q[H++];
if(H==MXQ) H=0;
cx=x%1000; x/=1000;
cy=x%1000; x/=1000;
cs=x;

for(i=0;i<4;i++)
{
nx=cx+dx[i]; ny=cy+dy[i]; ns=cs+1;
if(nx<0||nx>R||ny<0||ny>C||G[nx][ny]==-1||vis[nx][ny])
continue;
if(G[nx][ny]>0) // Got a new number, Update !
{
DigInd[nx][ny]=D;
DigsX[D]=nx; DigsY[D]=ny; DigsN[D]=G[nx][ny];
Dis[0][D]=ns;
Dis[D][0]=ns;
D++;
}
vis[nx][ny]=true;
Q[T++]=nx+1000*(ny+1000*ns);
if(T==MXQ) T=0;
}
}

// Distance to itself is 2 .. haha :)
for(i=0;i<D;i++) Dis[i][i]=2;
}

void bfsOnDig(int ci)
{
int sx=DigsX[ci], sy=DigsY[ci];
int i,j,k,cx,cy,cs,nx,ny,ns,x;

for(i=0;i<=R;i++) for(j=0;j<=C;j++) vis[i][j]=false;
int H=0,T=0;
Q[T++]=(sx+sy*1000);
vis[sx][sy]=true;

while(H!=T)
{
x=Q[H++];
if(H==MXQ) H=0;
cx=x%1000; x/=1000;
cy=x%1000; x/=1000;
cs=x;
for(i=0;i<4;i++)
{
nx=cx+dx[i]; ny=cy+dy[i]; ns=cs+1;
if(nx<0||nx>R||ny<0||ny>C||G[nx][ny]==-1||vis[nx][ny])
continue;
if(G[nx][ny]>0) Dis[ci][DigInd[nx][ny]]=ns;
vis[nx][ny]=true;
Q[T++]=(nx+1000*(ny+1000*ns));
if(T==MXQ) T=0;
}
}
}

int getMinSteps(int X)
{
int i,j,k,pn,cn,pi,ci,r=INF,mn;
for(i=0;i<D;i++) for(j=0;j<=X;j++) dp[i][j]=INF;
dp[0][0]=0;

for(i=1;i<=X;i++)
for(ci=1;ci<D;ci++)
{
pn=i-DigsN[ci];
if(pn<0) continue;
mn=INF;
for(pi=0;pi<D;pi++)
mn = MIN( mn , dp[pi][pn]+Dis[ci][pi]);
dp[ci][i]=mn;
}

for(i=0;i<D;i++) r = MIN( r , dp[i][X] );
return r;
}

int main()
{
int cases,i,j,k,x,sx,sy,s,res;
char ch,in[105];
scanf("%d",&cases);
while(cases--)
{
scanf("%d %d ",&R,&C);
for(i=0;i<=R;i++)
{
gets(in);
for(j=0;j<=C;j++)
{
ch = in[j];
if(ch=='*') sx=i,sy=j;
if(ch=='#') G[i][j]=-1;
else
{
if(ch>'0' && ch<='9') G[i][j]=ch-'0';
else G[i][j]=0;
}
}
}
scanf("%d",&x);

// find APSP ( = Every SSSP )
bfsStart(sx,sy);
for(i=1;i<D;i++) bfsOnDig(i);

res = getMinSteps(x);
if(res==INF) printf("-1\n");
else printf("%d\n",res);
}
return 0;
}

/* Code and Let Code ~ */``````

## Maze of Digits CodeChef Solution in JAVA

``````import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Main {

public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out = new PrintWriter(System.out);

for (int t = Integer.valueOf(in.readLine()); t > 0; t--) {
in.readLine();
String[] T = in.readLine().split(" ");
int n = Integer.valueOf(T[0]) + 1;
int m = Integer.valueOf(T[1]) + 1;

int[][] map = new int[n][m];
int sx = -1, sy = -1;

for (int i = 0; i < n; i++) {
String s = in.readLine();
for (int j = 0; j < m; j++) {
map[i][j] = s.charAt(j) == '#' ? -1 : s.charAt(j) >= '0' && s.charAt(j) <= '9' ? s.charAt(j) - '0' : 0;
if (s.charAt(j) == '*') {
sx = i;
sy = j;
}
}
}

int q = Integer.valueOf(in.readLine());
int res = -1;

int[][][] dp = new int[n][m][q+1];
start = finish = 0;
dp[qx[finish] = sx][qy[finish] = sy][qq[finish] = 0] = 1;
finish++;

while (finish - start > 0) {
int x = qx[start], y = qy[start], Q = qq[start], D = dp[x][y][Q];
start++;

if (Q == q) {
res = D - 1;
break;
}

for (int i = 0; i < dx.length; i++) {
int tox = x + dx[i], toy = y + dy[i];
if (tox >= 0 && tox < n && toy >= 0 && toy < m && map[tox][toy] != -1) {
int toq = Q + map[tox][toy];
if (toq <= q && dp[tox][toy][toq] == 0) {
dp[qx[finish] = tox][qy[finish] = toy][qq[finish] = toq] = D + 1;
finish++;
}
}
}
}

out.println(res);
}

out.flush();
}

static int[] qx = new int[101*101*101];
static int[] qy = new int[101*101*101];
static int[] qq = new int[101*101*101];
static int start, finish;

static int[] dx = new int[]{1, 0, -1, 0};
static int[] dy = new int[]{0, 1, 0, -1};
}

``````

## Maze of Digits CodeChef Solution in C#

``````using System;
using System.Collections;

public class K3
{
static int[] dr = { 0, 1, 0, -1 };
static int[] dc = { 1, 0, -1, 0 };

public static void Main(string[] args)
{
int t = int.Parse(Console.ReadLine());
while (t-- > 0) TestCase();
}

public static void TestCase()
{
Console.ReadLine();
string[] parts = Console.ReadLine().Split();
int M = 1 + int.Parse(parts[0]);
int N = 1 + int.Parse(parts[1]);
char[][] maze = new char[M][];
int sr = -1, sc = -1;
for (int i = 0; i < M; i++)
{
maze[i] = Console.ReadLine().ToCharArray();
for (int j = 0; j < N; j++)
{
if (maze[i][j] == '*')
{
sr = i;
sc = j;
maze[i][j] = '.';
}
}
}
int X = int.Parse(Console.ReadLine());

Queue Q = new Queue();
Q.Enqueue(new State(sr, sc, 0, X));
bool[, ,] done = new bool[M, N, X + 1];
while (Q.Count > 0)
{
State top = Q.Dequeue() as State;
if (done[top.r, top.c, top.x])
continue;
done[top.r, top.c, top.x] = true;
if (maze[top.r][top.c] == '#')
continue;
int nx = top.x;
if(maze[top.r][top.c] != '.') nx -= maze[top.r][top.c] - '0';
if (nx == 0)
{
Console.WriteLine(top.steps);
return;
}
for (int i = 0; i < 4; i++)
{
int nr = top.r + dr[i];
int nc = top.c + dc[i];
if (nr < 0 || nr >= M || nc < 0 || nc >= N || nx < 0 || done[nr, nc, nx]) continue;
Q.Enqueue(new State(nr, nc, top.steps + 1, nx));
}
}

Console.WriteLine("-1");
}

public class State
{
public int r, c, steps, x;

public State(int r, int c, int s, int x)
{
this.r = r;
this.c = c;
this.steps = s;
this.x = x;
}
}
}``````
##### Maze of Digits CodeChef Solution Review:

In our experience, we suggest you solve this Maze of Digits CodeChef Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Maze of Digits CodeChef Solution.

Find on CodeChef

##### Conclusion:

I hope this Maze of Digits CodeChef Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Programming Language in a business context; there are no prerequisites.

Keep Learning!

###### More Coding Solutions >>

Cognitive Class Answer