【hdu 5945】Fxx and game（递推|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): 347    Accepted Submission(s): 78

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

 

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【题解】【递推|dp+单调队列】

【从小往大处理：从1枚举到n，f[i]表示i最少经几次变换能变到1】

【f[i]=max(f[j](max(i-t,1)<=j<i),[k|i]f[i/k])+1，每次先判断当前数是否能整除k，再与f[j](max(i-t,1)<=j<i)中的每一个相比较，取最大值】

【那么，我们可以发现，实际就是在f[i/k]（i能整除k）和f[j](max(i-t,1)<=j<i)的最大值中选一个大的。那么，我们可以用单调队列来维护f[j](max(i-t,1)<=j<i)的最大值，这样就能省掉一重循环，就可以AC！】

[考试时，用dfs实现上述过程，结果栈溢出！]

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int t,k,n,T;
int f[1000010],que[1000010],minn[1000010];
int main()
{
//	freopen("int.txt","r",stdin);
//	freopen("my.txt","w",stdout);
	int i,j;
	scanf("%d",&T);
	while(T--)
	 {
	 	memset(f,127,sizeof(f));
	 	int head,tail; head=tail=1;
	 	f[1]=0; minn[tail]=0; que[tail]=1;
	 	scanf("%d%d%d",&n,&k,&t);
	 	for(i=2;i<=n;++i)
	     {
	     	if(!(i%k)) f[i]=f[i/k]+1;
	     	int m=max(i-t,1);
	     	while(tail>head&&que[head]<m) head++;
	     	f[i]=min(f[i],minn[head]+1);
	     	while(tail>=head&&f[i]<=minn[tail]) tail--;
			tail++; minn[tail]=f[i]; que[tail]=i; 
		 }
		printf("%d\n",f[n]);
	 }
	return 0;
 }