cf 424
Office Keys
首先显然有随人位置的递增，钥匙的位置也要递增，这样考虑两张做法：
1.$f(i,j)$ 表示前i个人，钥匙到第j个最少用的最大时间，然后$O(nK)$ dp
2.二分时间，对于每一个人选择当前能选择的最左面的钥匙 $O((n+K) logn)$
#include <bits/stdc++.h> #define LL long long const int N = 2010; using namespace std; int n,m,pre[N],a[N],b[N]; int Abs(int x) { if(x<0) return -x; return x; } int p; bool check(int tim) { int j = 1; for(int i=1;i<=n;i++) { while(j<=m && Abs(a[i]-b[j])+(LL)Abs(b[j]-p) > tim) j++; if(j>m) return 0; else j++; } return 1; } int main() { int maxv = 0; scanf("%d%d%d",&n,&m,&p); maxv = p; for(int i=1;i<=n;i++) scanf("%d",&a[i]), maxv = max(maxv ,a[i]); for(int i=1;i<=m;i++) scanf("%d",&b[i]), maxv = max(maxv, b[i]); sort(a+1,a+n+1); sort(b+1,b+m+1); int l=0, r=0; for(int i=1;i<=n;i++) l = max(l, Abs(p-a[i])); for(int i=1;i<=n;i++) r = max(r, Abs(b[i]-a[i]) + Abs(p-b[i])); while(r-l>5) { int mid = (l+(LL)r)>>1LL; if(check(mid)) r = mid; else l = mid; } for(int i=l;i<=r;i++) if(check(i)) { cout<<i<<endl; break; } return 0; }
Cards Sorting
方法1：直接用splay模拟。
方法2：考虑每次删掉当前最小数的代价是上一个最小数和它之间的循环dist，BIT或链表即可。
(取自CF)
#include <bits/stdc++.h> const int N = 100010; #define INF 0x3f3f3f3f #define LL long long using namespace std; struct node { node *ch[2]; int v, sum, minv; int cmp(int x) { if(ch[0]->sum + 1 == x) return -1; return x > ch[0]->sum; } node* init(int x); void update() { sum = ch[0]->sum + ch[1]->sum + 1; minv = min(v, min(ch[0]->minv, ch[1]->minv)); } }spT[N], Null, *root; int tot=0; node* node::init(int x) { v=x; minv=x; ch[0]=ch[1]=&Null; sum=1; return this; } node* build(int src[],int l,int r) { if(l==r) return spT[++tot].init(src[l]); int mid=(l+r)>>1; node* tmp=spT[++tot].init(src[mid]); if(l<mid) tmp->ch[0]=build(src,l,mid-1); if(mid<r) tmp->ch[1]=build(src,mid+1,r); tmp->update(); return tmp; } void rot(node* &o,int x) { node* tmp=o->ch[x]; o->ch[x]=tmp->ch[x^1]; tmp->ch[x^1]=o; o->update(); o=tmp; o->update(); } void splay(node* &o,int k) { int t=o->cmp(k); if(t==-1) return; if(t) k-=o->ch[0]->sum+1; int t1=o->ch[t]->cmp(k); if(~t1) { if(t1) k-=o->ch[t]->ch[0]->sum+1; splay(o->ch[t]->ch[t1],k); if(t1==t) rot(o,t); else rot(o->ch[t],t1); } rot(o,t); } int find_min(node *&o,int minv,int now) { if(o->ch[0]->minv == minv) return find_min(o->ch[0],minv,now); else { if(o->v == minv) return now+o->ch[0]->sum; return find_min(o->ch[1],minv,now+o->ch[0]->sum+1); } } int n,a[N]; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); Null.minv = INF; a[0]=INF; a[n+1]=INF; root = build(a,0,n+1); int len = n+2; LL ans = 0; for(int i=1;i<=n;i++) { int tmp = find_min(root,root->minv,0); ans += (LL)tmp; splay(root,1); splay(root->ch[1],tmp+1); node* tp = root->ch[1]->ch[0]; root->ch[1]->ch[0] = &Null; root->ch[1]->update(); root->update(); splay(tp,tp->sum); tp = tp->ch[0]; len--; splay(root,len-tp->sum); splay(root->ch[0],len-tp->sum-1); root->ch[0]->ch[1] = tp; root->ch[0]->update(); root->update(); } cout << ans << endl; return 0; }
Bamboo Partition
显然要满足的条件等价于
$\sum_{i=1}^n { [\frac{a_i + d - 1}{d}] \cdot d - a_i} \leq K$
$F(d) = \sum_{i=1}^n { [\frac{a_i - 1}{d}] + 1} \cdot d \leq K + \sum {a_i}$
固定 $\sum_{i=1}^n { [\frac{a_i - 1}{d}] + 1}$，从而 $F(d)$ 单调。
对于前者直接找出所有得数相同的 $d$ 的区间，逐一计算即可。
复杂度$O(n^2 \sqrt {a_i} )$
通过调参可以 $O(n \sqrt {a_{max} n})$
#include <bits/stdc++.h> #define LL long long const int N = 110; using namespace std; int n,tot,a[N]; LL K; vector<int> v; int main() { cin>>n>>K; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); K += (LL)a[i]; int j=1; int tmp = a[i]-1; if(tmp>0) { for(int i=1;i<=tmp;i=j+1) j = tmp/(tmp/i), v.push_back(i); v.push_back(tmp+1); } } v.push_back(1); sort(v.begin(),v.end()); v.resize(distance(v.begin(), unique(v.begin(), v.end()))); LL ans = 1, sum; for(int i=0;i<(int)v.size() - 1;i++) { int l = v[i], r = v[i+1]-1; sum = 0; for(int j=1;j<=n;j++) sum += (a[j]-1LL)/l + 1LL; LL tmp = min(K/sum, (LL)r); if(tmp>=l) ans = tmp; } LL l = v[v.size()-1]; LL tmp = K/n; if(tmp >= l) ans = tmp; cout << ans << endl; }
