Fxx and game hdu 5945 单调队列dp-CFANZ编程社区

Fxx and game hdu 5945 单调队列dp_#define

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
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#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 pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair 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;
}

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