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HDU5945 Fxx and game(单调队列)

题意:

给你三个数x,k,t(1e6),表示你在x每次可以减1-t或者可以整除k的时候除以k

问你到达1的最小步数

思路:

这次BC简直福利场。。过了题就上分

我是用单调队列维护的前k个值,然后被卡掉了- -

看了题解改成单调队列维护,当时就没想到啊(其实是以为不会被卡就没改而已)

/* ***********************************************
Author        :devil
************************************************ */
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <string>
#include <time.h>
#include <cmath>
#include <stdlib.h>
#define LL long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ou(a) printf("%d\n",a)
#define pb push_back
#define mkp make_pair
template<class T>inline void rd(T &x)
{
    char c=getchar();
    x=0;
    while(!isdigit(c))c=getchar();
    while(isdigit(c))
    {
        x=x*10+c-0;
        c=getchar();
    }
}
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
using namespace std;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=1e6+10;
pair<int,int> dp[N];
int f[N],t,x,k,v;
int main()
{
#ifndef ONLINE_JUDGE
//IN
#endif
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&x,&k,&v);
        int l=0,r=0;
        for(int i=2;i<=x;i++)
        {
            if(i<=v+1)
            {
                f[i]=1;
                dp[r].first=i,dp[r++].second=1;
                continue;
            }
            while(i-dp[l].first>v) l++;
            f[i]=dp[l].second+1;
            if(i%k==0) f[i]=min(f[i],f[i/k]+1);
            while(r>l&&f[i]<dp[r-1].second) r--;
            dp[r].first=i,dp[r++].second=f[i];
        }
        printf("%d\n",f[x]);
    }
    return 0;
}

 

HDU5945 Fxx and game(单调队列)