DFS与回溯法

DFS

排列数

问题: 生成1~n的排列
思路: 穷举所有可能 在生成结果数组前把重复的去掉
python code

A = [None for i in range(10)]N = 3def dfs(cur):    if cur == N:        print(A[:N])    else:        for i in range(1, N+1):            if i not in A[:cur]:                A[cur] = i                dfs(cur+1)dfs(0)

使用algorithm头文件的C++
C++ code

#include <iostream>#include <algorithm>using namespace std;int A[10];int N = 3;int main(){    for (int i = 0; i < N; i++)        A[i] = i+1;    do {        for (int i = 0; i < n; i++)            cout << A[i];        cout << endl;    } while(next_permutation(A, A+N));    return 0;}

素数环

问题: 数字1~n围成一个n个节点的环, 不允许数字重复, 任意2个相邻数字相加, 结果均为素数, 打印所有素数环的组合.
思路: 同排列数, 多了素数判断.
python code

A = [None for i in range(0, 10)]N = 6def is_prime(n):    for i in range(2, n//2+1):        if n%i == 0:            return False    return Truedef dfs(cur):    if cur == N:        if is_prime(A[0]+A[N-1]):            print(A[:N])    else:        for i in range(1, N+1):            if i not in A[:cur]:                if cur == 0 or is_prime(i+A[cur-1]):                    A[cur] = i                    dfs(cur+1)dfs(0)

八皇后

问题: 8×8的国际象棋棋盘上摆放八个皇后,使其不能互相攻击到,有多少种解.
思路: 结果数组存放每个棋子的x坐标, 直到没有冲突放完八个棋子.
解决冲突:
1. 不可能行冲突
2. 列冲突解决同排列数
解决斜线冲突, 利用y-x的特性找出关系.
3. cur-Q[cur] == j-Q[j]
4. cur+Q[cur] == j+Q[j]
python code

Q = [None for i in range(0, 8)]CNT = 0N = 8def dfs(cur):    if cur == N:        global CNT        CNT += 1    else:        for i in range(0, N):            Q[cur] = i            ok = True            for j in range(0, cur):                if Q[cur]==Q[j] or cur-Q[cur]==j-Q[j] or cur+Q[cur]==j+Q[j]:                    ok = False            if ok:                dfs(cur+1)dfs(0)print('ans = ' + str(CNT))

回溯法

探索到某一步发现原先选择达不到目标, 就退回一步重新选择.
效率比普通DFS高. 可以优化排列数和素数环的程序

访问过某个数后标记这个数已经访问再进行下一步的搜索
搜索完毕之后退回前一个搜索点, 再清除标记, 继续搜索.

排列数和素数环使用回溯法
python code

A = [None for i in range(0, 10)]V = [False for i in range(0, 10)]N = 4def dfs(cur):    if cur == N:        print(A[:N])    else:        for i in range(1, N+1):            if not V[i]:                V[i] = True                A[cur] = i                dfs(cur+1)                V[i] = Falsedfs(0)

python code

A = [None for i in range(0, 10)]V = [False for i in range(0, 10)]N = 6def is_prime(n):    for i in range(2, n//2+1):        if n%i == 0:            return False    return Truedef dfs(cur):    if cur == N:        if is_prime(A[0]+A[N-1]):            print(A[:N])    else:        for i in range(1, N+1):            if not V[i]:                if cur == 0 or is_prime(i+A[cur-1]):                    V[i] = True                    A[cur] = i                    dfs(cur+1)                    V[i] = Falsedfs(0)

八皇后的回溯法
在棋子放下的时候直接判断列和斜线是否存在其他皇后,并标记那些坐标
python code

Q = [None for i in range(8)]VM = [False for i in range(8)]VL = [False for i in range(16)]VR = [False for i in range(16)]CNT = 0N = 8def dfs(cur):    if cur == N:        global CNT        CNT += 1    else:        for i in range(0, N):            if not VM[i] and not VL[cur+i] and not VR[cur-i+N]:                VM[i] = VL[cur+i] = VR[cur-i+N] = True                Q[cur] = i                dfs(cur+1)                VM[i] = VL[cur+i] = VR[cur-i+N] = Falsedfs(0)print('ans = ' + str(CNT))

二维DFS搜索

二维搜索使用BFS的效率更高且附带最短路径, 用DFS编写扫雷核心算法.
核心算法: 点击到一个空白区域, 可以向四面八方展开空白块和数字.
DFS算法的代码更容易编写(BFS当然也可以完成)
python code

'''dfs(x-1, y)dfs(x+1, y)dfs(x, y-1)dfs(x, y+1)dfs(x-1, y+1)dfs(x+1, y-1)dfs(x-1, y-1)dfs(x+1, y+1)'''# serach 8 directionsdx = [-1,  1,  0,  0, -1,  1, -1,  1]dy = [ 0,  0, -1,  1,  1, -1, -1,  1]def pour(x, y):    if M[x][y] == 0:        show_empty(x, y)        for i in range(0, 8):            pour(x+dx[i], y+dy[i])    elif M[x][y] > 0:        show_num(x, y)