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 = 3

def 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 = 6

def is_prime(n):
    for i in range(2, n//2+1):
        if n%i == 0:
            return False
    return True

def 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 = 0
N = 8

def 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 = 4

def 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] = False

dfs(0)

python code

A = [None for i in range(0, 10)]
V = [False for i in range(0, 10)]
N = 6
def is_prime(n):
    for i in range(2, n//2+1):
        if n%i == 0:
            return False
    return True

def 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] = False

dfs(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 = 0
N = 8

def 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] = False


dfs(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 directions

dx = [-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)