HDOJ 1253 胜利大逃亡（bfs）

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 = ;
struct Point
{
    int a, b, c;
    int time;
    Point() { a = ; b = ; c = ; time = ; }
    Point(int x, int y, int z, int step_time)
    {
        a = x; b = y; c = z; time = step_time;
    }
};

int ans = ;
int A, B, C, game_times;
int map[MAX_N][MAX_N][MAX_N];
int move[][] =
{{, , -}, {, , }, {, -, }, {, , }, {-, , }, {, , }};
Point queue[];

int Bfs()
{
    int head, tail;
    Point temp;

    head = tail = ;
    temp.a = temp.b = temp.c = temp.time = ;
    queue[tail++] = temp;

    while (head != tail)
    {
        temp = queue[head++];
        for (int i = ; i < ; ++i)
        {
            int next_a = temp.a + move[i][];
            int next_b = temp.b + move[i][];
            int next_c = temp.c + move[i][];

            if (next_a <  || next_b <  || next_c <
                || next_a >= A || next_b >= B || next_c >= C
                || map[next_a][next_b][next_c] >  || temp.time +  > game_times)
                continue;

            if (next_a == A -  && next_b == B -  && next_c == C - )
                return temp.time + ;

            Point next(next_a, next_b, next_c, temp.time + );
            map[next_a][next_b][next_c] = next.time;
            queue[tail++] = next;
        }
    }
    return  -;
}

int main()
{
    int k;

    scanf("%d", &k);
    while (k--)
    {
        scanf("%d %d %d %d", &A, &B, &C, &game_times);
        for (int i = ; i < A; ++i)
        for (int j = ; j < B; ++j)
        for (int k = ; k < C; ++k)
            scanf("%d", &map[i][j][k]);

        printf("%d\n", Bfs());
    }
    return ;
}