Fxx and game hdu 5945 单调队列dp

dfs你怕是要爆炸

考虑dp;

很容易想到 dp[ i ] 表示到 i 时的最少转移步数；

那么： dp[ i ]= min( dp[ i ],dp[ i-j ]+1 );

其中 i-t<=j<=i;

当 i%k==0时 ,dp[ i ]=min( dp[ i ],dp[ i/k ]+1 )；

很明显这种要T到飞起；

我们要优化dp；

1e6的数据考虑O(n)级别的；

队列优化：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 2000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

int dp[maxn];
int q[maxn];

int main() {
//	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int T; cin >> T;
	while (T--) {
		ms(dp);
		int n, k, t; cin >> n >> k >> t;
		int l = 1, r = 1;
		dp[1] = 0;
		q[r++] = 1;
		for (int i = 2; i <= n; i++) {
			dp[i] = inf;
			while (l < r&&q[l] < i - t)l++;
			if (l < r)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-1]] >= dp[i])r--;
			q[r++] = i;
		}
		cout << dp[n] << endl;
	}
	return 0;
}