UVA 10047-the monocycle (BFS)

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Test instructions: From the starting point to the end point, each second can choose to go forward, left, right, each forward a wheel to the next color, a total of 5 colors, the beginning of green contact with the ground, north, requires the final green contact with the ground, to reach the target point and the shortest time.

Idea: Compared with the ordinary BFS, more than two additional conditions, so to the status of comprehensive, but also corresponding to add two dimensions. Water problem.

See the code for details:

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include < string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include < cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue># Define MAX (a) > (b)? ( A):(B) #define MIN (a) < (b) ( A):(B)) using namespace Std;typedef long Long ll;const double PI = ACOs ( -1.0); const double EPS = 1e-6;const int mod = 5;con St int INF = 1000000000;const int maxn = 50;int T,n,m,d[maxn][maxn][6][4],kase=0;int dir[] = {0,1,2,3};    Cardinal dx[] = {0, 1, 0, -1};int dy[] = {1, 0,-1, 0};struct node {int x, y, col, dir; Node (int x=0, int y=0, int col=0, int dir=0): x (x), Y (y), Col (Col), dir (dir) {}}s,t;char s[maxn][maxn];int BFS () {queue&lt    ;node> Q;    memset (d,-1, sizeof (d));    D[s.x][s.y][s.col][s.dir] = 0;    Q.push (S);        while (!q.empty ()) {node u = q.front (); Q.pop (); if (u.x= = T.x && u.y = = T.y && u.col = 0) return D[u.x][u.y][u.col][u.dir];        Node v = u;        v.x = v.x + Dx[v.dir];        V.Y = V.y + Dy[v.dir];        V.col = (v.col + 1)% MoD; if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m && D[v.x][v.y][v.col][v.dir] =            =-1 && s[v.x][v.y]! = ' # ') {D[v.x][v.y][v.col][v.dir] = D[u.x][u.y][u.col][u.dir] + 1;        Q.push (v);        } v = u;        V.dir = (v.dir-1 + 4)% 4; if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m && D[v.x][v.y][v.col][v.dir] =            =-1 && s[v.x][v.y]! = ' # ') {D[v.x][v.y][v.col][v.dir] = D[u.x][u.y][u.col][u.dir] + 1;        Q.push (v);        } v = u;        V.dir = (v.dir + 1)% 4; if (v.x >= 1 && v.x <= n && v.y >= 1 && v.y <= m && D[v.x][v.y][v.col][v.dir] =   =-1 && s[v.x][v.y]! = ' # ') {         D[v.x][v.y][v.col][v.dir] = D[u.x][u.y][u.col][u.dir] + 1;        Q.push (v); }} return-1;}        int main () {while (~scanf ("%d%d", &n,&m)) {if (n = = 0 && m = = 0) return 0;        if (Kase) printf ("\ n");        for (int i=1;i<=n;i++) {scanf ("%s", s[i]+1);  } for (int i=1;i<=n;i++) {for (int j=1;j<=m;j++) {if (s[i][j] = = ' s ') s = node (i, J,                0, 3);            else if (s[i][j] = = ' t ') t = node (i, j, 0, 0);        }} int ans = BFS ();        printf ("Case #%d\n", ++kase);        if (ans = =-1) printf ("Destination not reachable\n");    else printf ("Minimum time =%d sec\n", ans); } return 0;}

UVA 10047-the monocycle (BFS)