#2102：A计划（DFS和BFS剪枝搜索）

题意:

有几个比较坑的地方总结一下, 很容易误解:

1. 遇到#就必须走
2. #不消耗时间
3. #对面如果也是#也不能走, 要不然无限循环了
4. 最短路径剪枝时, 发现不能走的#是要把两步都标注为-1并跳出

题解:

一道经典的DFS+剪枝的题目, 这道题数据量过百了(大致范围), 直接裸的DFS+回溯是过不去滴, 必须要剪枝
可以看出, 当时间超过T不可达时直接跳出为可行性剪枝, 储存最短路径并只在当前小于最短路径才继续搜索为最优性剪枝

BFS

#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
int n, m, T;
typedef struct Node {
    int x, y, z;
    int t;
}Node;
int A[2][15][15];
int ans[2][15][15];	//存储状态
int vis[2][15][15];	//是否走过
int sx, sy, sz, ex, ey, ez;
int arr[4][2] = { {0,-1},{0,1},{1,0},{-1,0} };	//四个方向
queue<Node>q;
bool check(int x, int y, int z) {
    if (x >= 0 && x < 2 && y >= 0 && y < n&&z >= 0 && z < m) {
        //不能是墙，时空机后面不能是时空机，且没有走过
        if (A[x][y][z] && !vis[x][y][z] && A[x][y][z] != 2)return 1;
        else return 0;
    }
    return 0;
}
void update_ans(Node &s) {
    //更新状态
    if (ans[s.x][s.y][s.z] < 0 || s.t < ans[s.x][s.y][s.z])ans[s.x][s.y][s.z] = s.t;
}
void BFS() {
    Node start;
    start.x = sx; start.y = sy; start.z = sz;
    start.t = 0;
    vis[sx][sy][sz] = 1;
    q.push(start);
    while (!q.empty()) {
        Node v = q.front(); q.pop();
        int i;
        update_ans(v);
        if (ans[ex][ey][ez] >= 0)return;
        for (i = 0; i < 4; i++) {
            Node t = v;
            t.y += arr[i][0]; t.z += arr[i][1];
            t.t++;
            if (A[t.x][t.y][t.z] == 2) { t.x = (t.x + 1) % 2; }
            if (check(t.x, t.y, t.z)) {
                vis[t.x][t.y][t.z] = 1;
                q.push(t);
            }
        }
    }
}
int main() {
    int C;
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    scanf("%d", &C);
    while (C--) {
        scanf("%d%d%d", &n, &m, &T);
        int i, j, k;
        while (!q.empty())q.pop();
        memset(A, 0, sizeof(A));
        for (i = 0; i < 2; i++) {
            for (j = 0; j < n; j++) {
                for (k = 0; k < m; k++) {
                    char ch;
                    scanf("%c", &ch);
                    while (ch == '
' || ch == ' ' || ch == '' || ch == '?')scanf("%c", &ch);
                    if (ch == 'S') {
                        A[i][j][k] = 1; sx = i;
                        sy = j; sz = k;
                    }
                    else if (ch == 'P') {
                        A[i][j][k] = 9; ex = i;
                        ey = j; ez = k;
                    }
                    else if (ch == '*')A[i][j][k] = 0;
                    else if (ch == '#')A[i][j][k] = 2;
                    else if (ch == '.')A[i][j][k] = 1;
                }
            }
        }
        memset(ans, -1, sizeof(ans));
        memset(vis, 0, sizeof(vis));
        BFS();
        if (ans[ex][ey][ez] >= 0 && ans[ex][ey][ez] <= T)printf("YES
");
        else printf("NO
");
    }
    return 0;
}

DFS

#include<iostream>
#include<string.h>
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
int n, m, times, flag;
int dx[] = { 0,-1,0,1 };
int dy[] = { -1,0,1,0 };
bool book[2][11][11];
char map[2][11][11];
bool check(int x, int y) {
    if (x >= 0 && x < n && y >= 0 && y < m)return 1;
    else return 0;
}
bool dfs(int x, int y, int z, int t) {
    if (map[z][x][y] == 'P') {
        if (t <= times)return 1;
        else return 0;
    }
    for (int i = 0; i < 4; ++i) {
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (check(nx, ny)) {
            if (!book[1 - z][nx][ny] && map[z][nx][ny] == '#'&&map[1 - z][nx][ny] != '*'&&map[1 - z][nx][ny] != '#')
            {
                book[1 - z][nx][ny] = 1;
                if (dfs(nx, ny, 1 - z, t + 1))
                    return 1;
                book[1 - z][nx][ny] = 0;
            }
            else
                if (!book[z][nx][ny] && map[z][nx][ny] != '#'&&map[z][nx][ny] != '*')
                {
                    book[z][nx][ny] = 1;
                    if (dfs(nx, ny, z, t + 1))
                        return 1;
                    book[z][nx][ny] = 0;
                }
        }
        

    }
    return 0;
}
int main() {
    int t;
    cin >> t;
    while (t--) {
        cin >> n >> m >> times;
        ms(book, 0);
        ms(map, 0);
        for (int i = 0; i < 2; i++){
            getchar();
            for (int j = 0; j < n; j++){
                for (int k = 0; k < m; k++)
                    cin >> map[i][j][k];
                getchar();
            }
        }
        book[0][0][0] = 1;
        if (dfs(0, 0, 0, 0))cout << "YES" << endl;
        else cout << "NO" << endl;
        
    }
}