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=1000000007;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef __int64 ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1ll<<55;
const double eps=1e-8;
const double inf=~0u>>1;
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[1000006];
int dis[1000006];
int main()
{
        scanf("%d",&cas);
        for(int i=1; i<=cas; i++)
        {
            scanf("%d %d %d",&x,&k,&t);
            l=r=1;
            d[1]=1;
            for(int j=2;j<=x;j++) dis[j]=INF;
            dis[1]=0;
            for(int j=2; 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]]+1;
                if(j%k==0) dis[j]=min(dis[j],dis[j/k]+1);
                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
",dis[x]);
        }
    return 0;
}