hdu 5945 Fxx and game【dp+单调队列】

Fxx and game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 651    Accepted Submission(s): 155

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

29 2 111 3 3

 

Sample Output

43

 

题目大意：

一共有两种操作，问将x变成1需要的最少操作次数：

①X=X-I（0<=I<=tt）

②X=X/K（X%k==0）

思路：

1、考虑dp，设定dp【i】表示我们从数字1变成数字i使用的最少操作次数，那么不难推出其状态转移方程：

dp【i】=min（dp【i/k】+1，dp【i】）（i此时是k的倍数）

dp【i】=min（dp【i-j】+1，dp【i】）【0<=j<=tt】

2、根据数据范围可知，此时第二种状态转移方程的时间复杂度将会是：O（x*t），最大能达到10^12，是一个很大的操作数量级，那么考虑单调队列优化即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<iostream>using namespace std;int dp[1000500];int que[1000500];int pos[1000500];int main(){    int tt;    scanf("%d",&tt);    while(tt--)    {        int x,k,t;        scanf("%d%d%d",&x,&k,&t);        int head=1;        int tot=1;        que[head]=0;        pos[head]=1;        memset(dp,0x3f3f3f3f,sizeof(dp));        dp[1]=0;        for(int i=2;i<=x;i++)        {            if(i%k==0)            {                dp[i]=min(dp[i/k]+1,dp[i]);            }            while(head<=tot&&pos[head]<i-t)head++;            dp[i]=min(dp[i],que[head]+1);            while(head<=tot&&dp[i]<que[tot])tot--;            que[++tot]=dp[i],pos[tot]=i;        }        printf("%d\n",dp[x]);    }}