Susan

/**
Problema Turn - Constantinescu Eduard
Inspirata din problema Traseu - Carmen Minca
Algoritm de rezolvare: Lee cu mici modificari
Complexitate Timp O(n*n*n)
*/


#include <bits/stdc++.h>

using namespace std;

FILE *fin = fopen("turn.in", "r");
FILE *fout = fopen("turn.out", "w");

int n;

struct
{
    int prop, val;
    ///prop = 1 => scara sus
    ///prop = 2 => scara jos
    ///prop = 3 => groapa
    ///prop = 0 => gol
} cub[105][105][105];

struct
{
    int x, y, z; ///coordonatele in cub
} C[105*105*105], vec, poz, sf;

int dx[] = {1, 0, 0, 0, 0, -1}; ///cele 4 directii
int dy[] = {0, 0, -1, 0, 1, 0};
int dz[] = {0, 1, 0, -1, 0, 0};

/**


dx = 1 0 0 0 0 -1
dy = 0 0-1 0 1 0
dz = 0 1 0-1 0 0

x = sus
y = dreapta
z = spate

*/

void bordare() ///bordare simpla a matricei
{
    register int i, j;
    for(i = 0; i <= n+1; i++)
        for(j = 0; j <= n+1; j++)
        {
            cub[i][0][j].val = -1;
            cub[i][n+1][j].val = -1;
        }
    for(i = 0; i <= n+1; i++)
        for(j = 0; j <= n+1; j++)
        {
            cub[i][j][n+1].val = -1;
            cub[i][j][0].val = -1;
        }
    for(i = 0; i <= n+1; i++)
        for(j = 0; j <= n+1; j++)
        {
            cub[n+1][i][j].val = -1;
            cub[0][i][j].val = -1;
        }
}

void citire()
{
    int z, s, j, o;
    fscanf(fin, "%d%d%d%d%d", &n, &z, &s, &j, &o); ///citirea valorilor
    int a, b, c;
    for(int i = 1; i <= z; i++) /// se citesc coord ziduri
    {
        fscanf(fin, "%d%d%d", &a, &b, &c);
        cub[a][b][c].val = -1;
    }
    for(int i = 1; i <= s; i++)/// se citesc coord scari sus
    {
        fscanf(fin, "%d%d%d", &a, &b, &c);
        cub[a][b][c].prop = 1;
    }
    for(int i = 1; i <= j; i++)/// se citesc coord scari jos
    {
        fscanf(fin, "%d%d%d", &a, &b, &c);
        cub[a][b][c].prop = 2;
    }
    for(int i = 1; i <= o; i++)/// se citesc coord gropi
    {
        fscanf(fin, "%d%d%d", &a, &b, &c);
        cub[a][b][c].prop = 3;
    }

    fscanf(fin, "%d%d%d", &poz.x, &poz.y, &poz.z); ///se citesc coord
    fscanf(fin, "%d%d%d", &sf.x, &sf.y, &sf.z); /// de inceput si sf

}

void rezolvare() ///un lee cu mici modificari
{
    int prim, ultim, k, s = 1, l;

    prim = ultim = 0;
    C[0] = poz;
    cub[poz.x][poz.y][poz.z].val = 1; ///marcam pozitia de inceput

    while(prim <= ultim && cub[sf.x][sf.y][sf.z].val == 0) /// cat timp am el in coada
    {
        ///si nu am ajuns la sf
        poz = C[prim];
        ++prim; /// extrag din coada

        if(cub[poz.x][poz.y][poz.z].prop == 3) ///este groapa
        {
            int d = 0;
            vec = poz; /// pornim de la pozitia gropii
            while(vec.x > 1 && cub[vec.x][vec.y][vec.z].prop == 3)
                --vec.x, d++;///cat timp suntem in cub si suntem pe o groapa, vom cadea
            if(cub[vec.x][vec.y][vec.z].val == 0) ///este liber
            {
                cub[vec.x][vec.y][vec.z].val = cub[poz.x][poz.y][poz.z].val+d;
                ultim++;///marchez drumul si adaug in coada
                C[ultim] = vec;
            }
        }

        else
        {
            if(cub[poz.x][poz.y][poz.z].prop == 1)
                {
                    k = 0;
                    l = 4;
                }
            else if(cub[poz.x][poz.y][poz.z].prop == 2)
            {
                k = 1;
                l = 5;
            }
            else
            {
                k = 1;
                l = 4;
            }
            for(k; k <= l; k++) ///parcurg in cele 4 directii posibile
            {
                vec.x = poz.x + dx[k];
                vec.y = poz.y + dy[k];
                vec.z = poz.z + dz[k];
                if(cub[vec.x][vec.y][vec.z].val == 0)///este liber
                {
                    cub[vec.x][vec.y][vec.z].val = cub[poz.x][poz.y][poz.z].val + 1;
                    ultim++; ///marchez drumul si adaug in coada
                    C[ultim] = vec;
                }
            }

        }
    }
    fprintf(fout, "%d\n", cub[sf.x][sf.y][sf.z].val);
    ///afisez rezultatul
}

int main()
{
    citire();
    bordare();
    rezolvare();
    return 0;
}