Fxx and game(bc上的题)

Fxx and game

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 102

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.ifk|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
题意：
青年理论计算机科学家Fxx给的学生设计了一款数字游戏。
一开始你将会得到一个数X，每次游戏将给定两个参数k,t, 任意时刻你可以对你的数执行下面两个步骤之一：
1.X=X−i(1<=i<=t)。
2.若X为k的倍数，X=X/k。
现在Fxx想要你告诉他最少的运行步骤，使X变成1。
方法：
单调队列维护dp 或者 bfs剪枝
这里就说说bfs剪枝：
操作2是不需要剪枝的,因为它的下一状态就一个
于是着手点在操作1,X的下一状态可以是X-t~X-1中的任意一种,有点多,我们是不是可以省掉一些状态呢?
可以的,对于X-t~X-1这些状态中,我们从小往大遍历,当遇到某一状态已经访问过,那我们就不再取该状态及之后状态
为什么呢?理由很简单,我们在步数少的情况下已经到达该状态,那之后的状态还留着干嘛用呢
#include<stdio.h>#include<iostream>#include<queue>using namespace std;#define N 1000005struct node{    int x,c;}now,we;int v[N];int bfs(int x,int k,int t){    queue<node>q;    if(k==1&&t)    return (x+t-2)/t;    int i,Min;    node u;    for(i=1;i<=x;i++)        v[i]=0;        now.x=x;        now.c=0;    q.push(now);    v[x]=1;    while(!q.empty())    {        u=q.front();        q.pop();        if(u.x==1)            return u.c;        if(u.x%k==0&&!v[u.x/k])        {            we.x=u.x/k;            we.c=u.c+1;            q.push(we);            v[u.x/k]=1;        }        Min=min(t,u.x-1);        for(i=Min;i>=1;i--)            if(!v[u.x-i])            {                we.x=u.x-i;                we.c=u.c+1;                q.push(we);                v[u.x-i]=1;            }            else                break;//剪枝    }}int main(){    int T,X,k,t;    scanf("%d",&T);    while(T--)    {        scanf("%d%d%d",&X,&k,&t);        printf("%d\n",bfs(X,k,t));    }    return 0;}