HDU 2579 Dating with girls (2) BFS

/*---------------------------------------------time:14:45-15:25 2012.2.13 stratege:bfs (keypoint: Defines a three-dimensional array, nu
                M[X][Y][Z]; X, y is used to store the current maze coordinates, Z is used to store the time traveled and the modulus of k in the topic.     Because x, y can go back and forth, so we need to open one-dimensional array, update the state Acmer:johnsonddu problem:2579 (Dating with Girls (2)) Judge status:accepted runid:5111908 language:c++ author:a312745658------------------------------------- ----------*/#include <iostream> #include <stdio.h> #include <string.h> #include <math.h> #incl

Ude <memory.h> #include <queue> #include <algorithm> using namespace std;
    #define INF 0xfffffff struct Node {int x, y;
int step;
}n, M;
int direct[4][2] = {1, 0, 0,-1,-1, 0, 0, 1};
Char map[105][105];
BOOL Flag;
int num[105][105][11];
int R, C, K;
int SX, SY, DX, dy;

int tcase;
    int main () {int I, j, K;
    Cin >> Tcase; while (Tcase-) {cin >> R &Gt;> C >> K; 
                for (i = 0; i < R; i + +) for (j = 0; J < C; J + +) {cin >> map[i][j];
                if (map[i][j] = = ' Y ') sx = i, sy = j;
            if (map[i][j] = = ' G ') dx = i, dy = j; 
                    } for (i = 0; i < R; i + +) for (j = 0; J < C; J + +) for (k = 0; k < one; k + +)     Num[i][j][k] = INF;

        The number of steps to initialize is maximum to update the status flag = FALSE;
        n.x = SX;
        N.Y = sy;

        N.step = 0;
        Queue <Node> Q;

        Q.push (n); while (!
            Q.empty ()) {m = Q.front ();

            Q.pop ();
                if (m.x = = DX && m.y = = dy) {flag = true;
            break;
            }//cout << "(" << m.x << "," << m.y << ")" << Endl;
   for (i = 0; i < 4; i + +)         {n.x = m.x + direct[i][0];
                N.Y = M.y + direct[i][1];

                N.step = M.step + 1;
                    if (n.x >= 0 && n.y >= 0 && n.x < R && N.Y < C) {
                    if (map[n.x][n.y] = = ' # ' && n.step% k! = 0)//if n.step% K = = 0, the stone will disappear, can walk continue;
                        if (Num[n.x][n.y][n.step%k] <= n.step)//Update step number.
                    Continue;
                    Num[n.x][n.y][n.step%k] = N.step;
                Q.push (n);
        }}} if (flag) printf ("%d\n", m.step);

    else printf ("Please give me another chance!\n");
} return 0;
 }