Codeforces 877 problem D Olya and energy drinks BFS pruning

Source: Internet
Author: User

Transmission Door
Test instructions: Is given a starting point and an end point, each can go to four directions up to the K-step, each walk spent a second, asked to go to the end of the minimum number of seconds required.
Idea: Direct BFS will definitely t, we need to add a few pruning, one is when we go to some point, the point has been reached, and at this time than from the current point of time to go to this point is shorter, then do not go, because the next point from this point to a better point. See the code for specifics.

AC Code

const int MAXN = 1E3+5;
int Cas=1;
int dx[] = {1,-1,0,0};
int dy[] = {0,0,1,-1};
    void Solve () {int n,m,k;
    int Sx,sy,ex,ey;
        while (~SCANF ("%d%d%d", &n,&m,&k)) {for (int i=1;i<=n;i++) {scanf ("%s", mapp[i]+1);
        } scanf ("%d%d%d%d", &sx,&sy,&ex,&ey); Fill (vis,0);
        Fill (Dis,inf);
        Q.push (Make_pair (Sx,sy));
        Vis[sx][sy] = 1;
        Dis[sx][sy] = 0;
            while (!q.empty ()) {pair<int,int>tmp = Q.front ();
            Q.pop ();
                    for (int i=0;i<4;i++) {for (int j=1;j<=k;j++) {int xx =[i]*j;
                    int yy =[i]*j;
                    if (xx<1 | | xx>n | | yy<1 | | yy>m) break;
                    if (mapp[xx][yy] = = ' # ') break; if (Dis[xx][yy]!=inf && Dis[xx][yy] < DIs[][] + 1) break;
                    if (Vis[xx][yy]) continue;
                    VIS[XX][YY] = 1;
                    DIS[XX][YY] = dis[][] + 1;
                Q.push (Make_pair (XX,YY));
        }}} if (Dis[ex][ey]!=inf) cout << Dis[ex][ey] << Endl;
    else cout << "-1" << Endl; }
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