HDOJ 2102 A计划(bfs)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2102

思路分析：

<1>搜索方法分析：由于需要寻找最短的找到公主的路径，所以采用bfs搜索

<2>需要注意的地方：

1）如果当前状态为‘#‘字符，需要传送到另外一层，但是从该层到另外一层的时间是不会计算；

2）如果当前状态为‘#‘字符，则应该考虑该状态能否拓展；因为另一层的相同位置的字符可能为

‘S‘，‘P‘，‘.‘，‘*‘或者‘#‘，所以对于另一层相同位置的字符可能的各种情况应该考虑到并处理；

只有当字符为’P’或者’.’时才能被传送，其他字符不能传送，即该状态不能拓展，需要被剪枝；

PS:由于没有考虑到另一层的各种情况，导致WA多次，希望引以为鉴。

代码如下：

#include <iostream>
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;

const int MAX_N = 10 + 5;
char maze[2][MAX_N][MAX_N];
bool visited[2][MAX_N][MAX_N];
int N, M, T;
int move_[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

struct Node
{
    int x, y, z;
    int time;
    Node() { x = y = z = time = 0; }
    Node(int i, int j, int k, int t) { x = i; y = j; z = k; time = t; }
};

int Bfs()
{
    Node start;
    queue<Node> state;

    state.push(start);
    visited[0][0][0] = true;
    while (!state.empty())
    {
        int x, y, z, t;
        Node temp_state;

        temp_state = state.front();
        state.pop();
        x = temp_state.x;
        y = temp_state.y;
        z = temp_state.z;
        t = temp_state.time;
        for (int i = 0; i < 4; ++i)
        {
            char maze_ch = maze[z][x][y];
            int next_x = x + move_[i][0];
            int next_y = y + move_[i][1];
            int next_z = z;
            int next_time = t + 1;
            char maze_n_ch = maze[next_z][next_x][next_y];

            if (next_x < 0 || next_x >= N || next_y < 0 || next_y >= M
                || next_time > T || visited[next_z][next_x][next_y])
                continue;
            if (maze_n_ch == ‘#‘)
                next_z = 1 - next_z;
            maze_n_ch = maze[next_z][next_x][next_y];
            if (maze_n_ch == ‘*‘ || maze_ch == ‘#‘)
                continue;
            if (maze_n_ch == ‘P‘)
                return next_time;
            Node next_state(next_x, next_y, next_z, temp_state.time + 1);
            state.push(next_state);
            visited[next_z][next_x][next_y] = true;
        }
    }
    return -1;
}

int main()
{
    int case_times;

    scanf("%d", &case_times);
    while (case_times--)
    {
        int ans = 0;

        scanf("%d %d %d", &N, &M, &T);
        for (int i = 0; i < 2; ++i)
        for (int j = 0; j < N; ++j)
            scanf("%s", maze[i][j]);
        memset(visited, 0, sizeof(visited));
        ans = Bfs();
        if (ans == -1)
            printf("NO\n");
        else
            printf("YES\n");
    }
    return 0;
}