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深度优先搜索 DFS(Depath First Search, DFS)

作者:互联网

深度优先搜索是一种枚举所有完整路径以遍历所有情况的搜索方法。(不撞南墙不回头)

DFS一般用递归来实现,其伪代码思路过程一般如下:

void DFS(必要的参数){
    if (符和遍历到一条完整路径的尾部){
        更新某个全局变量的值
    }
    if (跳出循环的临界条件){
        return;
    }
    对所有可能出现的情况进行递归
}

 

常见题型1:

 代码实现:

 1 #include <stdio.h>
 2 const int maxn = 30;
 3 int n, V, maxVal = 0;        // 物品减数, 背包容量,最大价值maxValue
 4 int w[30];
 5 int c[30];
 6 int ans = 0;        // 最大价值
 7 
 8 // dfs, index是物品编号,nowW是当前所收纳的物品容量,nowC是当前所收纳的物品的总价值
 9 void dfs(int index, int nowW, int nowC){
10     if (index == n){
11         return;
12     }
13     dfs(index + 1, nowW, nowC);        // 不选第index件商品
14     if (nowW + w[index] <= V){        // 选第index件商品,但是先判断容量是否超限
15         if (nowC + c[index] > ans){
16             ans = nowC + c[index];        // 更新最大价值
17         }
18         dfs(index + 1, nowW + w[index], nowC + c[index]);
19     }
20 }
21 
22 int main()
23 {
24     scanf("%d %d", &n, &V);
25     for (int i = 0; i < n; i++){
26         scanf("%d", &w[i]);            // 每件物品的重量
27     }
28     for (int i = 0; i < n; i++){
29         scanf("%d", &c[i]);            // 每件物品的价值
30     }
31 
32     dfs(0, 0, 0);
33     printf("%d\n", ans);
34 
35     return 0;
36 }

 

常见题型二:

枚举从N 个整数找那个选择K个数(有时这个数可能可以重复)的所有方案(有时要打印这个方案的序列)

 代码实现:

 1 #include <stdio.h>
 2 #include <vector>
 3 using namespace std;
 4 
 5 const int maxn = 30;
 6 // 从包含n个数的序列A中选k个数使得和为x, 最大平方和为maxSumSqu;
 7 int n, k, sum, maxSumSqu = -1, A[maxn];
 8 vector<int> temp, ans;        // temp存放临时方案,ans存放平方和最大的方案
 9 
10 void DFS(int index, int nowK, int nowSum, int nowSumSqu){
11     if (nowK == k && nowSum == sum){
12         if (nowSumSqu > maxSumSqu){
13             maxSumSqu = nowSumSqu;
14             ans = temp;
15         }
16         return;
17     }
18     // 如果已经处理完n个数,或者选择了超过k个数,或者和超过x
19     if (index == n && nowK > k && nowSum > sum){
20         return;
21     }
22 
23     // 选这个数
24     temp.push_back(A[index]);
25     DFS(index + 1, nowK + 1, nowSum + A[index], nowSumSqu + A[index] * A[index]);
26     
27     // 不选这个数
28     // 先把刚加到temp中的数据去掉
29     temp.pop_back();
30     DFS(index + 1, nowK, nowSum, nowSumSqu);
31 }

如果选出的k个数可以重复,那么只需将上面“选这个数的 index + 1 改成 index 即可”

DFS(index + 1, nowK + 1, nowSum + A[index], nowSumSqu + A[index] * A[index]);

改成

DFS(index, nowK + 1, nowSum + A[index], nowSumSqu + A[index] * A[index]);

 

DFS题型实战:

          1103 Integer Factorization (30分)

The K−P factorization of a positive integer N is to write N as the sum of the P-th power of K positive integers. You are supposed to write a program to find the K−P factorization of N for any positive integers N, K and P.

Input Specification:

Each input file contains one test case which gives in a line the three positive integers N (≤400), K (≤N) and P (1<P≤7). The numbers in a line are separated by a space.

Output Specification:

For each case, if the solution exists, output in the format:

N = n[1]^P + ... n[K]^P

where n[i] (i = 1, ..., K) is the i-th factor. All the factors must be printed in non-increasing order.

Note: the solution may not be unique. For example, the 5-2 factorization of 169 has 9 solutions, such as 12​2​​+4​2​​+2​2​​+2​2​​+1​2​​, or 11​2​​+6​2​​+2​2​​+2​2​​+2​2​​, or more. You must output the one with the maximum sum of the factors. If there is a tie, the largest factor sequence must be chosen -- sequence { a​1​​,a​2​​,⋯,a​K​​ } is said to be larger than { b​1​​,b​2​​,⋯,b​K​​ } if there exists 1≤L≤K such that a​i​​=b​i​​ for i<L and a​L​​>b​L​​.

If there is no solution, simple output Impossible.

Sample Input 1:

169 5 2

Sample Output 1:

169 = 6^2 + 6^2 + 6^2 + 6^2 + 5^2

Sample Input 2:

169 167 3

Sample Output 2:

Impossible

代码实现:
 1 #include <stdio.h>
 2 #include <vector>
 3 #include <algorithm>
 4 #include <math.h>
 5 using namespace std;
 6 
 7 // 从1 - 20中选出k个数,数可重复,使得这些数的p次方的和刚好等于n, 求这些序列中和最大的那个序列
 8 
 9 int A[21];
10 int flag = 1;
11 int n, k, p, maxSum = -1;
12 vector<int> ans, temp, fac;        // ans 存放最终序列, temp存放临时序列
13 
14 // 快速幂
15 int power(int i){
16     /*if (p == 1 )
17         return i;
18     if ((p & 1) != 0)
19         return i * power(i, p - 1);
20     else
21     {
22         int temp = power(i, p / 2);
23         return temp * temp;
24     }*/
25 
26     int ans = 1;
27     for (int j = 0; j < p; j++){
28         ans *= i;
29     }
30     return ans;
31 }
32 
33 // 求出所有不大于n的p次幂
34 void init(){
35     int i = 0, temp = 0;
36     while (temp <= n){
37         fac.push_back(temp);
38         temp = power(++i);
39     }
40 }
41 
42 // DFS
43 void DFS(int index, int nowK, int sum, int squSum){
44     // 临界条件
45     if (squSum == n && nowK == k){
46         if (sum > maxSum){
47             maxSum = sum;
48             ans = temp;
49         }
50         
51         return;
52     }
53 
54     if (sum > n || nowK > k){
55         return;
56     }
57 
58     if (index >= 1){
59         // 遍历所有可能的情况
60         // 选当前数
61         temp.push_back(index);
62         DFS(index, nowK + 1, sum + index, squSum + fac[index]);
63 
64         // 不选当前数
65         temp.pop_back();
66         DFS(index - 1, nowK, sum, squSum);
67 
68         
69     }
70 }
71 
72 int main()
73 {
74     // 读取输入
75     // freopen("in.txt", "r", stdin);
76     scanf("%d %d %d", &n, &k, &p);
77 
78     // 初始化fac数组
79     init();
80 
81     // DFS寻找最合适的序列
82     DFS(fac.size() - 1, 0, 0, 0);
83     
84     // 输出
85     // 如果ans的size大于1则说明有结果
86     if (maxSum != -1){
87         // 排序
88         printf("%d = %d^%d", n, ans[0], p);
89         for (int i = 1; i < ans.size(); i++){
90             printf(" + %d^%d", ans[i], p);
91         }
92     }else
93         printf("Impossible");
94 
95     // fclose(stdin);
96     return 0;
97 }

这个实战题主要就是要先把 所有不超过 N 的 i ^p都算出来,要不然会超时

 

 

标签:index,Search,return,temp,int,DFS,Depath,ans
来源: https://blog.51cto.com/u_14201949/2832520