dfs题型一

代码：

 #include <iostream>
 #include <algorithm>
 #include <vector>
 using namespace std;

 const int maxn = ;

 //序列A中n各数选k个数使得和为x,最大平方和为maxSumSqu
 int n, k, x, maxSumSqu = -, A[maxn];

 vector<int> temp, ans;

 void dfs(int index, int nowK, int sum, int sumSqu){
     if (nowK == k && sum == x){            //如果找到k个数的和为x
         if (sumSqu > maxSumSqu){        //且找到的比当前更优
             maxSumSqu = sumSqu;            //更新最大平方和
             ans = temp;                    //更新最优方案
         }

         return;
     }

     //已经处理完n个数，或者超过k个数，或者和超过x, 返回
     if (index == n || nowK  > k || sum > x)
         return;

     //选index号数
     temp.push_back(A[index]);
     dfs(index + , nowK + , sum + A[index], sumSqu + A[index] * A[index]);
     temp.pop_back();

     //不选index号数
     dfs(index + , nowK, sum, sumSqu);
 }

 int main()
 {
     freopen("in.txt", "r", stdin);
     scanf("%d %d %d", &n, &k, &x);
     for (int i = ; i < n; i++){
         scanf("%d", &A[i]);
         printf("%d ", A[i]);
     }

     dfs(, , , );

     printf("%d\n", maxSumSqu);
     fclose(stdin);
     return ;
 }

如果每个元素可以重复被选，那么上面的程序只需改动一点点：将29行改为：

 dfs(index, nowK + 1, sum + A[index], sumSqu + A[index] * A[index]);

其实这题用K层迭代也能解决。

运用这个思路求解下面这道题：

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:

Impossibl

解答代码为：

 #include <stdio.h>
 #include <vector>
 #include <math.h>

 using namespace std;
 const int maxn = ;
 int fac[maxn];                //存储所有可能的项
 int n, k, p;
 int maxFacSum = -;
 vector<int> ans, temp;

 //算出fac数组
 int Fac(){
     int i = ;
     for (i = ; pow(i, p) <= n; i++){
         fac[i] = pow(i, p);
     }
     return i - ;
 }

 //index是fac的下标，nowK是当前已经叠加的项数，sum是当前之和，facSum是底数之和
 void dfs(int index, int nowK, int sum, int facSum){
     if (nowK == k && sum == n){
         if (facSum > maxFacSum){
             maxFacSum = facSum;
             ans = temp;
         }
         return;
     }
     //
     if (nowK > k || sum > n) return;

     if (index >= ){

         //选index项
         temp.push_back(index);
         dfs(index, nowK + , sum + fac[index], facSum + index);

         //不选index项
         temp.pop_back();
         dfs(index - , nowK, sum, facSum);
     }

 }

 int main()
 {
     //freopen("in.txt", "r", stdin);
     scanf("%d %d %d", &n, &k, &p);
     int index = Fac();
     dfs(index, , , );

     //如果ans的元素为空，则说明不存在有效解
     if (maxFacSum == -){
         printf("Impossible\n");
     }
     else{
         printf("%d = ", n);
         for (int i = ; i < k; i++){
             printf("%d^%d", ans[i], p);
             if (i < k - ){
                 printf(" + ");
             }
         }
     }

     //fclose(stdin);

     return ;
 }