HDU 5945 维护一个单调队列 dp

Fxx and game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 688    Accepted Submission(s): 162

Problem Description

Young theoretical computer scientist Fxx designed a game for his students.

In each game, you will get three integers X,k,t

.In each step, you can only do one of the following moves:

1.X=X−i(0<=i<=t)

.

2.

if k|X,X=X/k

.

Now Fxx wants you to tell him the minimum steps to make X

become 1.

 

Input

In the first line, there is an integer T(1≤T≤20)

indicating the number of test cases.

As for the following T

lines, each line contains three integers X,k,t(0≤t≤106,1≤X,k≤106)

For each text case,we assure that it's possible to make X

become 1。

 

Output

For each test case, output the answer.

 

Sample Input

2

9 2 1

11 3 3

 

Sample Output

4

3

 

Source

BestCoder Round #89

题意:给你三个数x，k，t 有两种操作 1.X=X−i(0<=i<=t) .2. if k|X,X=X/k . 问最少经过多少步使得x->1

题解:维护一个变换次数递增的单调队列

 /******************************
 code by drizzle
 blog: www.cnblogs.com/hsd-/
 ^ ^    ^ ^
  O      O
 ******************************/
 #include<bits/stdc++.h>
 #include<map>
 #include<set>
 #include<cmath>
 #include<queue>
 #include<bitset>
 #include<math.h>
 #include<vector>
 #include<string>
 #include<stdio.h>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #pragma comment(linker, "/STACK:102400000,102400000")
 using namespace std;
 #define  A first
 #define B second
 const int mod=;
 const int MOD1=;
 const int MOD2=;
 const double EPS=0.00000001;
 typedef __int64 ll;
 const ll MOD=;
 const int INF=;
 const ll MAX=1ll<<;
 const double eps=1e-;
 const double inf=~0u>>;
 const double pi=acos(-1.0);
 typedef double db;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 int cas;
 int x,k,t,l,r;
 int d[];
 int dis[];
 int main()
 {
         scanf("%d",&cas);
         for(int i=; i<=cas; i++)
         {
             scanf("%d %d %d",&x,&k,&t);
             l=r=;
             d[]=;
             for(int j=;j<=x;j++) dis[j]=INF;
             dis[]=;
             for(int j=; j<=x; j++)
             {
                 while(l<=r&&d[l]<j-t) l++;//把队列中没有用的点去掉
                 if(l<=r) dis[j]=dis[d[l]]+;
                 if(j%k==) dis[j]=min(dis[j],dis[j/k]+);
                 while(l<=r&&dis[d[r]]>=dis[j])r--;//将队列中变换次数多的去掉
                 d[++r]=j;
             }
             printf("%d\n",dis[x]);
         }
     return ;
 }