HDU 5945 Fxx and game (DP+单调队列)

题意：给定一个 x， k， t，你有两种操作，一种是 x - i (0 <= i <= t)，另一种是 x / k  （x % k == 0）。问你把x变成1需要的最少操作。

析：这肯定是DP，也想到可能是单调队列，但是不会啊。。。。就是胡搞了一发，虽然AC了，但是效率极低，比用单调队列少10倍。

dp[i] 表示把 i 变成 1，要用的最少步骤，然后每次取最优。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};
const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn];
int q[maxn];

int solve(int k, int x){
    int cnt = 0;
    while(x){
        x /= k;
        ++cnt;
    }
    return cnt-1;
}

int main(){
    int T;  cin >> T;
    while(T--){
        int x, k, t;
        scanf("%d %d %d", &x, &k, &t);
        if(k == 1){ printf("%d\n", (x-1)%t == 0 ? (x-1)/t : (x-1)/t+1);  continue; }
        else if(!t){ printf("%d\n", solve(k, x));   continue; }
        dp[1] = 0;  q[1] = 1;
        int l = 1, r = 1;
        for(int i = 2; i <= x; ++i){
            while(q[l] < i-t)  ++l;
            dp[i] = dp[q[l]] + 1;
            if(i % k == 0)  dp[i] = Min(dp[i], dp[i/k]+1);
            while(l <= r && dp[q[r]] >= dp[i])  --r;
            q[++r] = i;
        }
        printf("%d\n", dp[x]);
    }
    return 0;
}