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): 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

 

dp方程还是比较好推的，关键使用单调队列进行优化。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const LL MOD = 1e9 + 7;
const int MAXN = 1e6 + 10;
int dp[MAXN], que[MAXN], loc[MAXN];
int X, k, t;
int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
        scanf("%d%d%d", &X, &k, &t);
        dp[1] = 0; dp[0] = INF;
        int tail = 1, head = 1;
        que[head] = 0; loc[head] = 1;
        for (int i = 2; i <= X; i++) {
            dp[i] = INF;
            if (i%k == 0) dp[i] = dp[i/k] + 1;
            //找出队顶
            while (head <= tail && i - t > loc[head]) head++;
            dp[i] = min(dp[i], que[head] + 1);
            //新元素加入到队列
            while (head <= tail && dp[i] < que[tail]) tail--;
            que[++tail] = dp[i]; loc[tail] = i;
        }
        printf("%d\n", dp[X]);
    }
    return 0;
}