算法指南白書

維護一個四維數組，走一步更新一步

1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 7 const int INF = 1000000000; 8 const int maxr = 25 + 5; 9 const int maxc = 25 + 5; 10 int R, C, sr, sc, tr, tc; 11 char maze[maxr][maxc]; 12 13 struct State { 14 int r, c, dir, color; 15 State(int r, int c, int dir, int color):r(r),c(c),dir(dir),color(color) {} 16 }; 17 18 const int dr[] = {-1,0,1,0}; // north, west, south, east 19 const int dc[] = {0,-1,0,1}; 20 int d[maxr][maxc][4][5], vis[maxr][maxc][4][5];//橫坐標,縱坐標,方向,顏色 21 22 int ans; 23 queue<State> Q; 24 25 void update(int r, int c, int dir, int color, int v) { 26 if(r < 0 || r >= R || c < 0 || c >= C) return; // 不能走出邊界 27 if(maze[r][c] == '.' && !vis[r][c][dir][color]) { 28 Q.push(State(r, c, dir, color)); 29 vis[r][c][dir][color] = 1; 30 d[r][c][dir][color] = v; 31 if(r == tr && c == tc && color == 0) ans = min(ans, v); // 更新答案 32 } 33 } 34 35 void bfs(State st) { 36 d[st.r][st.c][st.dir][st.color] = 0; 37 vis[st.r][st.c][st.dir][st.color] = 1; 38 Q.push(st); 39 while(!Q.empty()) { 40 st = Q.front(); 41 Q.pop(); 42 int v = d[st.r][st.c][st.dir][st.color] + 1; 43 update(st.r, st.c, (st.dir+1)%4, st.color, v); // 左轉 44 update(st.r, st.c, (st.dir+3)%4, st.color, v); // 右轉 45 update(st.r+dr[st.dir], st.c+dc[st.dir], st.dir, (st.color+1)%5, v); // 前進 46 } 47 } 48 49 int main() { 50 int kase = 0; 51 while(scanf("%d%d", &R, &C) == 2 && R && C) { 52 for(int i = 0; i < R; i++) { 53 scanf("%s", maze[i]); 54 for(int j = 0; j < C; j++) 55 if(maze[i][j] == 'S') { 56 sr = i; 57 sc = j; 58 } else if(maze[i][j] == 'T') { 59 tr = i; 60 tc = j; 61 } 62 } 63 maze[sr][sc] = maze[tr][tc] = '.'; 64 ans = INF; 65 memset(vis, 0, sizeof(vis)); 66 bfs(State(sr, sc, 0, 0)); 67 68 if(kase > 0) printf("\n"); 69 printf("Case #%d\n", ++kase); 70 if(ans == INF) printf("destination not reachable\n"); 71 else printf("minimum time = %d sec\n", ans); 72 } 73 } View Code

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轉載于:https://www.cnblogs.com/ITUPC/p/5312976.html

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