【23.33%】【hdu 5945】Fxx and game

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 360 Accepted Submission(s): 84

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

【题解】

f[i] = min(f[i],f[i/k]+1)；
f[i] = min(f[i],f[i-t..i-1]+1);
第二行那个转移可用一个单调队列优化;
因为f[i+1]>=f[i];
dl的最左边维护的是i-t..i-1这个区间内f值最小的点的下标;然后往右f值依次递增;
如果新加入的点i,f[i]小于这个dl最右边x对应值f[x],则把x挤掉(那些被挤掉的元素是肯定没有用了的，如果到了某个时刻f[i]被挤掉了，因为i>x,则x肯定也要被挤掉)，重复上述步骤;
就能维护一个单调队列;
有点厉害.

#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson L,m,rt<<1
#define rson m+1,R,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 2e6;

const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;

int x,k,t,l,r;
int f[MAXN],dl[MAXN];

void input_LL(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)) t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void input_int(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)) t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int main()
{
    //freopen("F:\rush.txt", "r", stdin);
    int T;
    input_int(T);
    while (T--)
    {
        memset(f,INF,sizeof(f));
        input_int(x);input_int(k);input_int(t);
        l = 1;r = 1;dl[r] = 1;
        f[1] = 0;
        for (int i = 2;i <= x;i++)
        {
            while (l<=r && dl[l]<i-t)l++;
            if ((i%k)==0)
                f[i] = min(f[i],f[i/k]+1);
            if (l <=r)
                f[i] = min(f[i],f[dl[l]]+1);
            while (l<=r && f[i]<f[dl[r]]) r--;
            dl[++r] = i;
        }
        printf("%d
",f[x]);
    }
    return 0;
}