AOJ——分治递归之Exhaustive Search穷尽搜索

AOJ的分治递归的第一题：

Exhaustive Search

Write a program which reads a sequence A of n elements and an integer M, and outputs "yes" if you can make M by adding elements in A, otherwise "no". You can use an element only once.

You are given the sequence A and q questions where each question contains Mi.

Input

In the first line n is given. In the second line, n integers are given. In the third line q is given. Then, in the fourth line, q integers (Mi) are given.

Output

For each question Mi, print yes or no.

Constraints

n ≤ 20

q ≤ 200

1 ≤ elements in A ≤ 2000

1 ≤ Mi ≤ 2000

Sample Input 1
5
1 5 7 10 21
8
2 4 17 8 22 21 100 35
Sample Output 1
no
no
yes
yes
yes
yes
no
no

很明显的分治递归的思路，一开始没有看notes，搞了两天还是A不了，主要是穷举不完全，还有递归的终止条件比较模糊。

看一下notes：

Notes

You can solve this problem by a Burte Force approach. Suppose solve(p, t) is a function which checkes whether you can make t by selecting elements after p-th element (inclusive). Then you can recursively call the following functions:

solve(0, M)
solve(1, M-{sum created from elements before 1st element})
solve(2, M-{sum created from elements before 2nd element})
...

The recursive function has two choices: you selected p-th element and not. So, you can check solve(p+1, t-A[p]) and solve(p+1, t) in solve(p, t) to check the all combinations.

For example, the following figure shows that 8 can be made by A[0] + A[2].

一看就懂了，原来是这样确定递归条件啊，通过不断减去已列举的情况，再以最后的结果是否为零判断递归终止，其实想到了就非常简单（我还是太菜了）

看我的codes:

#include <iostream>
using namespace std;
int n;bool slove(int *A,int k, int m){if(m==0)return true;if(m<0||k==n)return false;return slove(A,k+1,m-A[k])||slove(A, k+1, m);}int main(){//    int n;cin>>n;int A[n];for(int i=0;i<n;i++)cin>>A[i];int q;cin>>q;int M[q];for(int i=0;i<q;i++){cin>>M[i];if((slove(A, 0, M[i]))==true)cout<<"yes"<<endl;elsecout<<"no"<<endl;}return 0;
}

总结：还是分解问题不够彻底，用其他的分解方法总是无法将这个问题分解完全。

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AOJ——分治递归之Exhaustive Search穷尽搜索

Exhaustive Search

Input

Output

Constraints

Sample Input 1

Sample Output 1

Notes

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