dfs题型一

代码：

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

const int maxn = 10000;

//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 = -1, 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 + 1, nowK + 1, sum + A[index], sumSqu + A[index] * A[index]);
    temp.pop_back();

    //不选index号数
    dfs(index + 1, nowK, sum, sumSqu);
}


int main()
{
    freopen("in.txt", "r", stdin);
    scanf("%d %d %d", &n, &k, &x);
    for (int i = 0; i < n; i++){
        scanf("%d", &A[i]);
        printf("%d ", A[i]);
    }

    dfs(0, 0, 0, 0);

    printf("%d
", maxSumSqu);
    fclose(stdin);
    return 0;
}

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 = 25;
int fac[maxn];                //存储所有可能的项
int n, k, p;
int maxFacSum = -1;
vector<int> ans, temp;

//算出fac数组
int Fac(){
    int i = 0;
    for (i = 0; pow(i, p) <= n; i++){
        fac[i] = pow(i, p);
    }
    return i - 1;
}

//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 >= 1){

        //选index项
        temp.push_back(index);
        dfs(index, nowK + 1, sum + fac[index], facSum + index);

        //不选index项
        temp.pop_back();
        dfs(index - 1, nowK, sum, facSum);
    }

}

int main()
{
    //freopen("in.txt", "r", stdin);
    scanf("%d %d %d", &n, &k, &p);
    int index = Fac();
    dfs(index, 0, 0, 0);
    
    //如果ans的元素为空，则说明不存在有效解
    if (maxFacSum == -1){
        printf("Impossible
");
    }
    else{
        printf("%d = ", n);
        for (int i = 0; i < k; i++){
            printf("%d^%d", ans[i], p);
            if (i < k - 1){
                printf(" + ");
            }
        }
    }

    //fclose(stdin);

    return 0;
}