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dfs你怕是要爆炸

考慮dp;

很容易想到 dp[ i ] 表示到 i 時的最少轉移步數；

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

其中 i-t<=j<=i;

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

很明顯這種要T到飛起；

我們要優化dp；

1e6的數據考慮O(n)級別的；

隊列優化：

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#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; }

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轉載于:https://www.cnblogs.com/zxyqzy/p/10311992.html

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