华为笔试题 2022.3.30_华为自动化控制工程师笔试

1、业务部署芯片

思路，就硬模拟

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {
    int m, n;
    cin >> m;
    cin >> n;

    char arr[n];
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }
    // int m = 5, n = 6;
    // char arr[n] = "ABABAA";

    //以上是输入  m 是芯片数，n是业务数，arr就是一个字符数组，放了一串ABABABAB
    char temp;
    int b[m];
    for (int i = 0; i < m; ++i) {
        b[i] = 0;
    }
    int a1 = 0;
    int index1 = 1;  //未满状态的第一块芯片编号
    int index2 = 1;  //全空状态的第一块芯片编号
    for (int i = 0; i < n - 1; ++i) {
        temp = arr[i];
        if (temp == 'A') {
            if (a1 + 1 == 4) {
                a1 = 0;
                index1 = index2;
            } else {
                a1++;
                if (index1 == index2) {
                    index2++;
                }
            }

        } else if (temp == 'B') {
            if (index1 == index2) {
                index1++;
                index2++;
            } else {
                index2++;
            }
        }
    }
    // 判断最后一个
    if (arr[n - 1] == 'A') {
        if (index1 > m) {
            cout << 0 << endl;
            cout << 0 << endl;
        } else {
            cout << index1 << endl;
            cout << a1 + 1 << endl;
        }

    } else {
        if (index2 > m) {
            cout << 0 << endl;
            cout << 0 << endl;
        } else {
            cout << index2 << endl;
            cout << 1 << endl;
        }
    }
    return 0;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72

第二题，标准的二维地图（带障碍物）两点最短距离

上dfs

#include <algorithm>
#include <iostream>
#include <string>

using namespace std;
//全局变量
const int MAX_X = 100;
const int MAX_Y = 100;
int min_step = 10000;
int number = 0;
int my_x, my_y;
int map[MAX_X][MAX_Y];
int book[MAX_X][MAX_Y];
int start_x, start_y;
int dest_x, dest_y;

void dfs(int x, int y, int step) {
    /*up, right, down, left*/
    int next[4][2] = {{0, -1}, {1, 0}, {0, 1}, {-1, 0}};
    int tx, ty;

    if (x == dest_x && y == dest_y) {
        if (step == min_step) {
            number++;
        }
        if (step < min_step) {
            min_step = step;
            number = 1;
        }

        return;
    }

    for (int i = 0; i < 4; i++) {
        tx = x + next[i][0];
        ty = y + next[i][1];
        if (tx > my_x || ty > my_y || tx < 0 || ty < 0)
            continue;

        if (map[tx][ty] == 0 && book[tx][ty] == 0) {
            book[tx][ty] = 1;
            dfs(tx, ty, step + 1);
            book[tx][ty] = 0;
        }
    }
}

int main() {
    cin >> my_x >> my_y;
    cin >> start_x >> start_y;
    cin >> dest_x >> dest_y;

    int num;
    cin >> num;

    //建一张二维表,初始化权为0
    int x, y;                        //湖泊的位置
    for (int i = 0; i < num; ++i) {  //高山湖泊用1代表
        cin >> x >> y;
        map[x][y] = 1;
    }

    book[start_x][start_y] = 1;
    dfs(start_x, start_y, 0);

    cout << number << " " << min_step << endl;
    return 0;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68

第三题，最大（深）的重复子树

652. 寻找重复的子树 差不多的题，只要把652的结果里面最长的那个返回就行了

万万没想到他给了一个层序遍历的数组，我不知道怎么变成一棵树，裂开，没做出来