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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1253
思路分析：因为问题需要寻找到达终点的最短的距离(最短的步数)，即在状态转换图上需要找出层次最浅的的状态(A-1, B-1, C-1)，
所以采用bfs更快能找出答案；另外，若采用dfs则比较困难，需要遍历所有的状态才能找出结果。
代码如下：
#include <cstdio> #include <iostream> const int MAX_N = 60; struct Point { int a, b, c; int time; Point() { a = 0; b = 0; c = 0; time = 0; } Point(int x, int y, int z, int step_time) { a = x; b = y; c = z; time = step_time; } }; int ans = 0; int A, B, C, game_times; int map[MAX_N][MAX_N][MAX_N]; int move[6][3] = {{0, 0, -1}, {0, 0, 1}, {0, -1, 0}, {0, 1, 0}, {-1, 0, 0}, {1, 0, 0}}; Point queue[1000000]; int Bfs() { int head, tail; Point temp; head = tail = 0; temp.a = temp.b = temp.c = temp.time = 0; queue[tail++] = temp; while (head != tail) { temp = queue[head++]; for (int i = 0; i < 6; ++i) { int next_a = temp.a + move[i][0]; int next_b = temp.b + move[i][1]; int next_c = temp.c + move[i][2]; if (next_a < 0 || next_b < 0 || next_c < 0 || next_a >= A || next_b >= B || next_c >= C || map[next_a][next_b][next_c] > 0 || temp.time + 1 > game_times) continue; if (next_a == A - 1 && next_b == B - 1 && next_c == C - 1) return temp.time + 1; Point next(next_a, next_b, next_c, temp.time + 1); map[next_a][next_b][next_c] = next.time; queue[tail++] = next; } } return -1; } int main() { int k; scanf("%d", &k); while (k--) { scanf("%d %d %d %d", &A, &B, &C, &game_times); for (int i = 0; i < A; ++i) for (int j = 0; j < B; ++j) for (int k = 0; k < C; ++k) scanf("%d", &map[i][j][k]); printf("%d\n", Bfs()); } return 0; }
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原文地址：http://www.cnblogs.com/tallisHe/p/4469888.html
