AcWing 172.立体推箱子_acwing192 立体推箱子-优快云博客

本质上就是bfs

难在状态的表示，转移，以及记录

这里说说我的方法吧

总所周知，如果状态表示成一个点，那么对于转移和记录来说是最简便的，而立体的箱子实际上只有三种状态，横着躺，竖着躺，和立起来，对于躺着的状态，只要记录最左上角的正方形方块的坐标即可，而立着的就是记录下面的那个方块的坐标，然后加个state变量来记录状态

代码如下：

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

const int MAXN = 510;
int N, M, ans;
char G[MAXN][MAXN];
bool vis[MAXN][MAXN][3];
struct Node{
    int x, y, state, d;
    Node():d(0){}
    Node(int x, int y, int state, int d):x(x), y(y), state(state), d(d){}
}bgn, edd;

void init(void){
    memset(vis, 0, sizeof(vis));
    memset(G, 0, sizeof(G));
    ans = 0;
    for(int i = 1; i <= N; i++){
        getchar();
        for(int j = 1; j <= M; j++) G[i][j] = getchar();
    }
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            if(G[i][j] == 'O'){
                edd.x = i; edd.y = j; edd.state = 0;
            }
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++)
            if(G[i][j] == 'X'){
                bgn.x = i; bgn.y = j;
                if(G[i][j + 1] == 'X') bgn.state = 2;
                else if(G[i + 1][j] == 'X') bgn.state = 1;
                else bgn.state = 0;
                return;
            }
}
bool is_illegal(int x, int y){
    if(x <= 0 || y <= 0 || x > N || y > M || G[x][y] == '#') return true;
    return false;
}
bool is_illegal(const Node &tt){
    if(vis[tt.x][tt.y][tt.state]) return true;
    if(is_illegal(tt.x, tt.y)) return true;
    if(tt.state == 0) return G[tt.x][tt.y] == 'E';
    else if(tt.state == 1) return is_illegal(tt.x + 1, tt.y);
    else return is_illegal(tt.x, tt.y + 1);
}
bool is_reach(const Node &tt){
    return tt.x == edd.x && tt.y == edd.y && tt.state == edd.state;
}
inline void bfs(void){
    queue<Node> q; q.push(bgn);
    while(!q.empty()){
        Node u = q.front(); q.pop();
        if(is_illegal(u)) continue;
        if(is_reach(u)){ans = u.d; break;}
        vis[u.x][u.y][u.state] = true;
        if(u.state == 0){
            q.push(Node(u.x - 2, u.y, 1, u.d + 1));
            q.push(Node(u.x + 1, u.y, 1, u.d + 1));
            q.push(Node(u.x, u.y - 2, 2, u.d + 1));
            q.push(Node(u.x, u.y + 1, 2, u.d + 1));
        }
        else if(u.state == 1){
            q.push(Node(u.x - 1, u.y, 0, u.d + 1));
            q.push(Node(u.x + 2, u.y, 0, u.d + 1));
            q.push(Node(u.x, u.y - 1, 1, u.d + 1));
            q.push(Node(u.x, u.y + 1, 1, u.d + 1));
        }
        else{
            q.push(Node(u.x - 1, u.y, 2, u.d + 1));
            q.push(Node(u.x + 1, u.y, 2, u.d + 1));
            q.push(Node(u.x, u.y - 1, 0, u.d + 1));
            q.push(Node(u.x, u.y + 2, 0, u.d + 1));
        }
    }
}

int main(void)
{
    while(scanf("%d%d", &N, &M) == 2 && N){
        init();
        bfs();
        if(ans == 0) printf("Impossible\n");
        else printf("%d\n", ans);
    }
}

回头看看题解的办法