hdu 5339 Untitled【dfs过】

最新推荐文章于 2021-10-21 18:26:48 发布

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本文探讨了一个关于整数序列的问题，目标是最小化选择的序列长度使得特定操作后的结果为零。介绍了问题背景、输入输出格式及样例，并通过三种不同方法进行了详细解析，最终采用排序加深度优先搜索的方法实现了高效求解。

Untitled

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 751 Accepted Submission(s): 403

Problem Description

There is an integer

a and

n integers

b1,…,bn . After selecting some numbers from

b1,…,bn in any order, say

c1,…,cr , we want to make sure that

a mod c1 mod c2 mod… mod cr=0 (i.e.,

a will become the remainder divided by

ci each time, and at the end, we want

a to become

0 ). Please determine the minimum value of

r . If the goal cannot be achieved, print

−1 instead.

Input

The first line contains one integer

T≤5 , which represents the number of testcases.

For each testcase, there are two lines:

1. The first line contains two integers

n and

a (

1≤n≤20,1≤a≤106 ).

2. The second line contains

n integers

b1,…,bn (

∀1≤i≤n,1≤bi≤106 ).

Output

T answers in

T lines.

Sample Input

Sample Output

  
   2
-1

Source

BestCoder Round #49 ($)

题目大意：找到最少的数字满足这么个条件：a mod c1 mod c2 mod… mod cr=0、其实就是找c的和是a的倍数，这个东西谁看都明白，然后给你n个数，让你去找最少的数字和、这里我直接想的就是搜索去做，但是捏~用什么方法很重要，这里我还是怕深搜超时、没有剪枝的话是很可能超时的，不如用bfs来的实惠、所以第一发写的bfs、实力MLE、、、、、、代码交给大家参考，仅供参考、、反正过不了：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define ll long long int
struct date
{
    ll val;
    int vis[21];
    int cont;
}now,nex;
int n;
ll a;
ll b[21];
void bfs()
{
    queue<date>s;
    for(int i=0;i<n;i++)
    {
        now.val=b[i];
        for(int j=0;j<n;j++)
        {
            now.vis[j]=0;
        }
        now.vis[i]=1;
        now.cont=1;
        s.push(now);
    }
    while(!s.empty())
    {
        now=s.front();
        s.pop();
        if(now.val%a==0)
        {
            printf("%d\n",now.cont);
            return ;
        }
        for(int i=0;i<n;i++)
        {
            nex=now;
            if(now.vis[i]==0)
            {
                nex.val=now.val+b[i];
                nex.cont=now.cont+1;
                nex.vis[i]=1;
                s.push(nex);
            }
        }
    }
    printf("-1\n");
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%I64d",&n,&a);
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&b[i]);
        }
        bfs();

    }
}

然后给代码修了修，无果，转战DFS、想到了排序，但是仅仅是先闪烁了一下这个思路，还是无脑dfs了一发，实力TLE、、、真是个令人悲伤的故事、这里也上代码，交给大家参考、、、

a mod c1 mod c2 mod… mod cr=0a mod c1 mod c2 mod… mod cr=0

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define ll long long int
int n;
ll a;
ll b[21];
int vis[21];
int output;
void dfs(ll val,int cont)
{
    if(val%a==0&&val!=0)
    {
        output=min(output,cont);
        return ;
    }
    ll x;
    for(int i=0;i<n;i++)
    {
        if(vis[i]==0)
        {
            x=val+b[i];
            vis[i]=1;
            dfs(x,cont+1);
            vis[i]=0;
        }
    }
    return ;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%I64d",&n,&a);
        for(int i=0;i<n;i++)
        {
            scanf("%I64d",&b[i]);
        }
        memset(vis,0,sizeof(vis));
        output=0x1f1f1f1f;
        dfs(0,0);
        if(output!=0x1f1f1f1f)
        printf("%d\n",output);
        else
        {
            printf("-1\n");
        }
    }
}

剪枝无果，最后上排序，30+ms就过了、、、

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
int output;
int b[21];
int n,mubiao;
void dfs(int val,int cont,int ans)
{
    if(val==0)
    {
        output=min(cont,ans);
        return;
    }
    if(cont==n)
    return ;
    dfs(val%b[cont],cont+1,ans+1);
    dfs(val,cont+1,ans);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        output=10000;
        scanf("%d%d",&n,&mubiao);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&b[i]);
        }
        sort(b,b+n);
        reverse(b,b+n);
        dfs(mubiao,0,0);
        if(output==10000)
        {
            printf("-1\n");
        }
        else
        {
            printf("%d\n",output);
        }
    }
}