DFS-简单递归
【简单递归】
题目描述
已知 n 个整数b1,b2,…,bn
以及一个整数 k（k＜n）。
从 n 个整数中任选 k 个整数相加，可分别得到一系列的和。
例如当 n=4，k＝3，4 个整数分别为 3，7，12，19 时，可得全部的组合与它们的和为：
3＋7＋12=22 3＋7＋19＝29 7＋12＋19＝38 3＋12＋19＝34。
现在，要求你计算出和为素数共有多少种。
例如上例，只有一种的和为素数：3＋7＋19＝29。
输入
第一行两个整数：n , k （1<=n<=20，k＜n）
第二行n个整数：x1,x2，…,xn （1<=xi<=5000000）
输出
一个整数（满足条件的方案数）。
样例输入
4 3
3 7 12 19
样例输出
1
代码
#include <iostream>
#include <cstring>
#include <vector>
using namespace std;
//判断是否是素数
bool isprime(int n)
{
if (n<1)return false;
for (int i=2;i*i<n;i++)
{
if (n%i==0)//n&(i-1)==0
return false;
}
return true;
}
int a[22],p[22], b[22];
bool vis[22];
int n, k, sum, ans;
int cnt = 0;
vector<vector<int> > index_vec;
vector<int> sum_vec;
void dfs(int index)
{
cout << "Enter dfs, " << ++cnt << "th" << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
if (index==k+1)
{
if (isprime(sum))
{
ans++;
cout << "sum:" << sum << " is prime." << endl;
}
//打印每一个元素组合
vector<int> v;
for (int i=1;i<=index-1;i++)
{
cout << p[i] << " ";
v.push_back(p[i]);
}
index_vec.push_back(v);
sum_vec.push_back(sum);
cout << endl;
cout << "sum:" << sum << endl;
return;
}
//遍历n个数
for (int i=1;i<=n;i++)
{
//条件1：该元素在本次遍历中是否被访问过
//条件2：保障遍历的i下标大于已统计过的最后一个元素下标
cout << "i:" << i << ", vis:" << vis[i] << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
if (vis[i]==false && i>p[index-1])
{
p[index] = i;//数组p用来保存元素下标，index为当前统计的元素总数目
vis[i] = true;//标记该元素已被该次计算统计过
sum += a[i];//更新和,在全排列过程中，对结果进行统计
cout << "Enter the next dfs" << endl;
cout << "sum:" << sum << ", i:" << i << ", vis[i]:" << vis[i] << endl;
dfs(index+1);//开始下一次递归计算
cout << "return from dfs" << endl;
//回溯
vis[i] = false;
sum -= a[i];
cout << "sum:" << sum << ", i:" << i << ", vis[i]:" << vis[i] << ", index:" << index << ", p[index-1]:" << p[index-1] << endl;
}
}
}
int main()
{
//初始化全局变量
memset(b, 0, sizeof(b));
memset(vis, false, sizeof(vis));
//获取输入
cin >> n >> k;
cout << "n=" << n << " ,k=" << k << endl;
for (int i=1; i<=n;i++)
{
cin >> a[i];
p[i]=i;
}
ans = 0;
dfs(1);
cout << ans << endl;
int seq=0;
for (auto it: sum_vec)
{
cout << "sum: " << it << endl;
for (auto it_idx: index_vec[seq])
{
cout << it_idx << " ";
}
cout << endl;
++seq;
}
return 0;
}
题目解析
循环中调用递归
代码打印帮助理解递归过程
4
3
n=4 ,k=3
3
7
12
19
Enter dfs, 1th, index:1, p[index-1]:0
i:1, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:3, i:1, vis[i]:1
Enter dfs, 2th, index:2, p[index-1]:1
i:1, vis:1, index:2, p[index-1]:1
i:2, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:10, i:2, vis[i]:1
Enter dfs, 3th, index:3, p[index-1]:2
i:1, vis:1, index:3, p[index-1]:2
i:2, vis:1, index:3, p[index-1]:2
i:3, vis:0, index:3, p[index-1]:2
Enter the next dfs
sum:22, i:3, vis[i]:1
Enter dfs, 4th, index:4, p[index-1]:3
1 2 3
sum:22
return from dfs
sum:10, i:3, vis[i]:0, index:3, p[index-1]:2
i:4, vis:0, index:3, p[index-1]:2
Enter the next dfs
sum:29, i:4, vis[i]:1
Enter dfs, 5th, index:4, p[index-1]:4
sum:29 is prime.
1 2 4
sum:29
return from dfs
sum:10, i:4, vis[i]:0, index:3, p[index-1]:2
return from dfs
sum:3, i:2, vis[i]:0, index:2, p[index-1]:1
i:3, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:15, i:3, vis[i]:1
Enter dfs, 6th, index:3, p[index-1]:3
i:1, vis:1, index:3, p[index-1]:3
i:2, vis:0, index:3, p[index-1]:3
i:3, vis:1, index:3, p[index-1]:3
i:4, vis:0, index:3, p[index-1]:3
Enter the next dfs
sum:34, i:4, vis[i]:1
Enter dfs, 7th, index:4, p[index-1]:4
1 3 4
sum:34
return from dfs
sum:15, i:4, vis[i]:0, index:3, p[index-1]:3
return from dfs
sum:3, i:3, vis[i]:0, index:2, p[index-1]:1
i:4, vis:0, index:2, p[index-1]:1
Enter the next dfs
sum:22, i:4, vis[i]:1
Enter dfs, 8th, index:3, p[index-1]:4
i:1, vis:1, index:3, p[index-1]:4
i:2, vis:0, index:3, p[index-1]:4
i:3, vis:0, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:3, i:4, vis[i]:0, index:2, p[index-1]:1
return from dfs
sum:0, i:1, vis[i]:0, index:1, p[index-1]:0
i:2, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:7, i:2, vis[i]:1
Enter dfs, 9th, index:2, p[index-1]:2
i:1, vis:0, index:2, p[index-1]:2
i:2, vis:1, index:2, p[index-1]:2
i:3, vis:0, index:2, p[index-1]:2
Enter the next dfs
sum:19, i:3, vis[i]:1
Enter dfs, 10th, index:3, p[index-1]:3
i:1, vis:0, index:3, p[index-1]:3
i:2, vis:1, index:3, p[index-1]:3
i:3, vis:1, index:3, p[index-1]:3
i:4, vis:0, index:3, p[index-1]:3
Enter the next dfs
sum:38, i:4, vis[i]:1
Enter dfs, 11th, index:4, p[index-1]:4
2 3 4
sum:38
return from dfs
sum:19, i:4, vis[i]:0, index:3, p[index-1]:3
return from dfs
sum:7, i:3, vis[i]:0, index:2, p[index-1]:2
i:4, vis:0, index:2, p[index-1]:2
Enter the next dfs
sum:26, i:4, vis[i]:1
Enter dfs, 12th, index:3, p[index-1]:4
i:1, vis:0, index:3, p[index-1]:4
i:2, vis:1, index:3, p[index-1]:4
i:3, vis:0, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:7, i:4, vis[i]:0, index:2, p[index-1]:2
return from dfs
sum:0, i:2, vis[i]:0, index:1, p[index-1]:0
i:3, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:12, i:3, vis[i]:1
Enter dfs, 13th, index:2, p[index-1]:3
i:1, vis:0, index:2, p[index-1]:3
i:2, vis:0, index:2, p[index-1]:3
i:3, vis:1, index:2, p[index-1]:3
i:4, vis:0, index:2, p[index-1]:3
Enter the next dfs
sum:31, i:4, vis[i]:1
Enter dfs, 14th, index:3, p[index-1]:4
i:1, vis:0, index:3, p[index-1]:4
i:2, vis:0, index:3, p[index-1]:4
i:3, vis:1, index:3, p[index-1]:4
i:4, vis:1, index:3, p[index-1]:4
return from dfs
sum:12, i:4, vis[i]:0, index:2, p[index-1]:3
return from dfs
sum:0, i:3, vis[i]:0, index:1, p[index-1]:0
i:4, vis:0, index:1, p[index-1]:0
Enter the next dfs
sum:19, i:4, vis[i]:1
Enter dfs, 15th, index:2, p[index-1]:4
i:1, vis:0, index:2, p[index-1]:4
i:2, vis:0, index:2, p[index-1]:4
i:3, vis:0, index:2, p[index-1]:4
i:4, vis:1, index:2, p[index-1]:4
return from dfs
sum:0, i:4, vis[i]:0, index:1, p[index-1]:0
1
sum: 22
1 2 3
sum: 29
1 2 4
sum: 34
1 3 4
sum: 38
2 3 4
Process finished with exit code 0
