Codeforces Round #659 (Div. 2) B1. Koa and the Beach (Easy Version)

Posted 2020-12-18 acmllf

 

 

 

 题意 小明从一岸游泳到另一岸，每片区域有水深，一旦水深超过L，小明就会淹死

同时每段时刻有海浪和退潮  

搜索一下 然后记忆化一下 

老了，搜索写半天 

#include<bits/stdc++.h>
using namespace std;
/*int main()
{
   // freopen("data2.in", "r", stdin);
   // freopen("data2.out", "w", stdout);
    int n,m;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        int x;
        cin>>x;
        a[i]=a[i-1]+x;
    }

    cin>>m;
    while(m--)
    {

        int l,r;
        cin>>l>>r;
        if(l==r)
            cout<<a[l]-a[l-1]<<endl;
        else
            cout<<a[r]-a[l-1]<<endl;
    }
}*/
int d[105];
 int n,k,l;
 int vis[105][205];
int dfs(int x,int t)
{
    if(vis[x][t])
    return 0;
    vis[x][t]=1;
    int flag=0;
    if(x==n+1)
    return 1;
    t=t%(2*k);
    int ad;
    if(t<=k)
        ad=t;
    else
        ad=k-(t-k);
    if(x!=0&&(d[x]+ad)>l) return 0;
        flag=max(flag,dfs(x+1,t+1));
    t++;
    t=t%(2*k);
    if(t<=k)
        ad=t;
    else
        ad=k-(t-k);
    if(x>0)
    {
       if(d[x]+ad>l)
       return max(flag,0);
    }
    flag=max(dfs(x,t),flag);
    return flag;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {   memset(vis,0,sizeof(vis));
        int flag=0;
        cin>>n>>k>>l;
        for(int i=1;i<=n;i++) {
                cin>>d[i];
                if(d[i]>l)
                 flag=1;
        }
        if(flag){cout<<"NO"<<endl;continue;}
        if(dfs(0,0))cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
}
/*int main()
{
    // freopen("data.in", "r", stdin);
        //freopen("test.in", "w", stdout);
    int t;
    cin>>t;
    mt19937 u32Rnd(time(0));
    cout<<t<<endl;
    while(t--)
    {

      int n;
      while((n=u32Rnd())<1);
      cout<<n<<endl;
    }
}*/