HDU 2579/bfs/dating with girls (2)

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/* Test instructions is a traditional maze plus a condition that the walls will disappear at the full multiples of k, so the shortest time to reach the exit is obtained.
The key point is that a point is taken up to K times, labeled Vis[x][y][time%k]. */#include <queue> #include <cstdio> #include <cstring> #include <iostream> #include <
Algorithm> using namespace std;
const int maxn=100+5;
int vis[maxn][maxn][15];
int dir[4][2]={0,1,1,0,0,-1,-1,0};
Char MAZE[MAXN][MAXN];
int Sx,sy,ex,ey;
int n,m,k;
    void Init () {memset (vis,0,sizeof (Vis));} struct node {int x;
    int y;
int time;
};
    int BFS () {node now;
    NOW.X=SX;
    Now.y=sy;
    now.time=0;
    queue<node>que;
    Que.push (now);
        while (!que.empty ()) {Now=que.front ();
        Que.pop ();
        if (Now.x==ex&&now.y==ey) return now.time;
            for (int i=0;i<4;i++) {int x=now.x+dir[i][0];
            int y=now.y+dir[i][1];
            int t=now.time+1;
            int mod=t%k; if (x>=1&&x<=n&&y>=1&&y<=m) {if (maze[x]
                    [y]!= ' # ' &&vis[x][y][mod]==0) {vis[x][y][mod]=t;
                    Node next= (node) {x,y,t};
                Que.push (next);
                    } else if (maze[x][y]== ' # ' &&mod==0&&vis[x][y][mod]==0) {
                    vis[x][y][mod]=t;
                    Node next= (node) {x,y,t};
                Que.push (next);
}}}} return-1;
    } int main () {int t;scanf ("%d", &t);
        while (t--) {scanf ("%d%d%d", &n,&m,&k);
        Init ();
            for (int i=1;i<=n;i++) {scanf ("%s", maze[i]+1);
                for (int j=1;j<=m;j++) if (maze[i][j]== ' Y ') sx=i,sy=j;
        else if (maze[i][j]== ' G ') ex=i,ey=j;
        } int Ans=bfs ();
        if (ans==-1) printf ("Please give me another chance!\n"); else PrinTF ("%d\n", ans);
} return 0; }