eoj 3279 爱狗狗的两个dalao(dfs)

来源：互联网发布：单机游戏编程语言编辑：程序博客网时间：2024/05/16 05:29

当时第一次见这题的时候我不在现场,没听到题解,今天看到有人过,突然想起,然后闲的蛋疼,过来写一下,1A.
一开始很智障,想着贪心,但是其实这样没有意义,类比背包.
N<18,直接搜索咯.
好像有许多地方可以优化,有一组数据会被卡的很惨,但是懒得想咯,我们又菜又懒的咸鱼是这样的.
一如既往丑陋低效的搜索.

/*  xzppp  */#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <string>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define PB push_backtypedef long long  LL;typedef unsigned long long ULL;typedef pair<int,int > pii;const int MAXN = 10000+17;const int MAXM = 20;const int MAXV = 1001+17;const int INF = 0x7fffffff;const int MOD = 1e9+7;int a[18+17],n,m,ans = INF;void dfs(int p,int num,vector<int > all){    if(num>=ans) return;    for (int i = 0; i < all.size(); ++i)    {        if(all[i]>=a[p])        {            all[i]-=a[p];            if(p==n-1)            {                ans = min(ans,num);                return;            }            else                dfs(p+1,num,all);            all[i]+=a[p];        }    }    all.push_back(m-a[p]);    if(p==n-1)    {        ans = min(ans,num+1);        return;    }    else        dfs(p+1,num+1,all);}int main(){    #ifndef ONLINE_JUDGE     FFF    #endif    cin>>n>>m;    for (int i = 0; i < n; ++i)        cin>>a[i];    vector<int > v;    dfs(0,0,v);    cout<<ans<<endl;    return 0;}

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