NBUT1479：How many(DFS)

最新推荐文章于 2019-04-25 19:50:38 发布

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最新推荐文章于 2019-04-25 19:50:38 发布

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分类专栏：搜索文章标签： DFS

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本文介绍了一种通过深度优先搜索（DFS）解决特定条件下的组合数问题的方法。问题要求从给定的不重复整数中找出所有可能的组合，使得这些整数的和等于指定的目标值，并统计这样的组合有多少种。文章提供了完整的代码实现，包括输入处理、递归搜索过程以及去重策略。

http://ac.nbutoj.com/Problem/view.xhtml?id=1479

问题描述
There are N numbers, no repeat. All numbers is between 1 and 120, and N is no more than 60. then given a number K(1 <= K <= 100). Your task is to find out some given numbers which sum equals to K, and just tell me how many answers totally,it also means find out hwo many combinations at most.
输入
There are T test cases.
For each case:
First line there is a number N, means there are N numbers.
Second line there is a number K, means sum is K.
Third line there lists N numbers.
See range details in describe.
输出
Output the number of combinations in total.
样例输入
```
2
5
4
1,2,3,4,5
5
6
1,2,3,4,5
```
样例输出

2
3

题意：给出n个数，要找出加起来等于k的方法有几种，要注意，1+2 = 3 与2+1=3，这是一样的情况

思路：与杭电SUM IT UP那道题是一样的，果断DFS

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

int a[1000],n,m,cnt;

int cmp(int a,int b)
{
    return a>b;
}

void dfs(int pos,int sum)
{
    int i;
    if(sum>m)
        return ;
    if(sum == m)
        cnt++;
    for(i = pos; i<n; i++)
    {
        dfs(i+1,sum+a[i]);
        while(i+1<n && a[i] == a[i+1])
            i++;
    }
}

int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i = 0; i<n-1; i++)
            scanf("%d,",&a[i]);
        scanf("%d",&a[i]);
        sort(a,a+n,cmp);
        cnt = 0;
        dfs(0,0);
        printf("%d\n",cnt);
    }

    return 0;
}