Codeforces 487B. Strip(求区间最值＋线段树上的dp)

Alexandra has a paper strip with n numbers on it. Let‘s call them ai from left to right.

Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:

Each piece should contain at least l numbers.
The difference between the maximal and the minimal number on the piece should be at most s.

Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input

The first line contains three space-separated integers n,?s,?l (1?≤?n?≤?105,?0?≤?s?≤?109,?1?≤?l?≤?105).

The second line contains n integers ai separated by spaces (?-?109?≤?ai?≤?109).

Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Sample test(s)

input
7 2 2
1 3 1 2 4 1 2

output
3

input
7 2 2
1 100 1 100 1 100 1

output
-1

Note

For the first sample, we can split the strip into 3 pieces: [1,?3,?1],?[2,?4],?[1,?2].

For the second sample, we can‘t let 1 and 100 be on the same piece, so no solution exists.

for(int i=1;i<=n;i++){ if(从i开始到n这段区间元素个数<L) continue; if(dp[i]==无穷大) continue; int to=i+L-1; if(maxto[i]<to) continue; for(int j=to+1;j<=maxto[i]+1;j++){ dp[j]=min(dp[j],dp[i]+1); } }

//线段树dp: #define MAXN 100100 #define INT (0x3f3f3f3f)*2 struct Segment_Tree{ int left,right; int dp,lazy; }tree[4*MAXN]; void plant_tree(int id,int l,int r){ tree[id].left=l,tree[id].right=r; tree[id].lazy=INT; if(l==r){ if(l==1) tree[id].dp=0; else tree[id].dp=INT; return; } int mid=(l+r)>>1; plant_tree(id<<1,l,mid); plant_tree((id<<1)+1,mid+1,r); tree[id].dp=min(tree[id<<1].dp,tree[(id<<1)+1].dp); } void push_down(int id){//懒操作pushdown if(tree[id].lazy==INT) return; tree[id].dp=min(tree[id].dp,tree[id].lazy); if(tree[id].left==tree[id].right){ tree[id].lazy=INT; return; } tree[id<<1].lazy=min(tree[id<<1].lazy,tree[id].lazy); tree[(id<<1)+1].lazy=min(tree[(id<<1)+1].lazy,tree[id].lazy); tree[id].lazy=INT; } void update_leaf(int id,int l,int r,int val){ if(tree[id].left==l && tree[id].right==r){ tree[id].lazy=min(tree[id].lazy,val); push_down(id); return; } push_down(id); int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) update_leaf(id<<1,l,r,val); else if(mid<l) update_leaf((id<<1)+1,l,r,val); else{ update_leaf(id<<1,l,mid,val); update_leaf((id<<1)+1,mid+1,r,val); } } int query(int id,int l,int r){ if(tree[id].left==l && tree[id].right==r){ push_down(id); return tree[id].dp; } push_down(id); int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) return query(id<<1,l,r); else if(mid<l) return query((id<<1)+1,l,r); else return min(query(id<<1,l,mid),query((id<<1)+1,mid+1,r)); }

//Hello. I'm Peter. #include<cstdio> #include<iostream> #include<sstream> #include<cstring> #include<string> #include<cmath> #include<cstdlib> #include<algorithm> #include<functional> #include<cctype> #include<ctime> #include<stack> #include<queue> #include<vector> #include<set> #include<map> using namespace std; typedef long long ll; typedef long double ld; #define peter cout<<"i am peter"<<endl #define input freopen("data.txt","r",stdin) #define randin srand((unsigned int)time(NULL)) #define INT (0x3f3f3f3f)*2 #define LL (0x3f3f3f3f3f3f3f3f)*2 #define gsize(a) (int)a.size() #define len(a) (int)strlen(a) #define slen(s) (int)s.length() #define pb(a) push_back(a) #define clr(a) memset(a,0,sizeof(a)) #define clr_minus1(a) memset(a,-1,sizeof(a)) #define clr_INT(a) memset(a,INT,sizeof(a)) #define clr_true(a) memset(a,true,sizeof(a)) #define clr_false(a) memset(a,false,sizeof(a)) #define clr_queue(q) while(!q.empty()) q.pop() #define clr_stack(s) while(!s.empty()) s.pop() #define rep(i, a, b) for (int i = a; i < b; i++) #define dep(i, a, b) for (int i = a; i > b; i--) #define repin(i, a, b) for (int i = a; i <= b; i++) #define depin(i, a, b) for (int i = a; i >= b; i--) #define pi 3.1415926535898 #define eps 1e-6 #define MOD 1000000007 #define MAXN 100100 #define N #define M //线段树求最值 struct Segment_TreeforRMQ{ int left,right; int min,max; }treeRMQ[4*MAXN]; void plant_treeforRMQ(int id,int l,int r,Segment_TreeforRMQ* tree,int *a){ tree[id].left=l,tree[id].right=r; if(l==r){ tree[id].min=tree[id].max=a[l]; return; } int mid=(l+r)>>1; plant_treeforRMQ(id<<1,l,mid,tree,a); plant_treeforRMQ((id<<1)+1,mid+1,r,tree,a); tree[id].min=min(tree[id<<1].min,tree[(id<<1)+1].min); tree[id].max=max(tree[id<<1].max,tree[(id<<1)+1].max); } int query_min(int id,int l,int r,Segment_TreeforRMQ* tree){ if(tree[id].left==l && tree[id].right==r){ return tree[id].min; } int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) return query_min(id<<1,l,r,tree); else if(mid<l) return query_min((id<<1)+1,l,r,tree); else return min(query_min(id<<1,l,mid,tree),query_min((id<<1)+1,mid+1,r,tree)); } int query_max(int id,int l,int r,Segment_TreeforRMQ* tree){ if(tree[id].left==l && tree[id].right==r){ return tree[id].max; } int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) return query_max(id<<1,l,r,tree); else if(mid<l) return query_max((id<<1)+1,l,r,tree); else return max(query_max(id<<1,l,mid,tree),query_max((id<<1)+1,mid+1,r,tree)); } //查询最大值-最小值 int maxsubmin(int from,int to){ return query_max(1,from,to,treeRMQ)-query_min(1,from,to,treeRMQ); } //线段树dp: struct Segment_Tree{ int left,right; int dp,lazy; }tree[4*MAXN]; void plant_tree(int id,int l,int r){ tree[id].left=l,tree[id].right=r; tree[id].lazy=INT; if(l==r){ if(l==1) tree[id].dp=0; else tree[id].dp=INT; return; } int mid=(l+r)>>1; plant_tree(id<<1,l,mid); plant_tree((id<<1)+1,mid+1,r); tree[id].dp=min(tree[id<<1].dp,tree[(id<<1)+1].dp); } void push_down(int id){//懒操作pushdown if(tree[id].lazy==INT) return; tree[id].dp=min(tree[id].dp,tree[id].lazy); if(tree[id].left==tree[id].right){ tree[id].lazy=INT; return; } tree[id<<1].lazy=min(tree[id<<1].lazy,tree[id].lazy); tree[(id<<1)+1].lazy=min(tree[(id<<1)+1].lazy,tree[id].lazy); tree[id].lazy=INT; } void update_leaf(int id,int l,int r,int val){ if(tree[id].left==l && tree[id].right==r){ tree[id].lazy=min(tree[id].lazy,val); push_down(id); return; } push_down(id); int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) update_leaf(id<<1,l,r,val); else if(mid<l) update_leaf((id<<1)+1,l,r,val); else{ update_leaf(id<<1,l,mid,val); update_leaf((id<<1)+1,mid+1,r,val); } } int query(int id,int l,int r){ if(tree[id].left==l && tree[id].right==r){ push_down(id); return tree[id].dp; } push_down(id); int mid=(tree[id].left+tree[id].right)>>1; if(r<=mid) return query(id<<1,l,r); else if(mid<l) return query((id<<1)+1,l,r); else return min(query(id<<1,l,mid),query((id<<1)+1,mid+1,r)); } void printno(){ printf("%d\n",-1); exit(0); } int n,s,l; int a[MAXN],maxto[MAXN]; int main() { cin>>n>>s>>l; repin(i,1,n){ scanf("%d",a+i); } if(n<l) printno();//当n<l的时候，不成立 plant_treeforRMQ(1,1,n,treeRMQ,a);//建立树，为了求a数组的最值 int to=1; repin(i,1,n){//求出只满足maxi-mini<=s的条件下，每个点i最远可往右到哪个点 to=max(to,i); while(1){ if(to==n){ maxto[i]=to; break; } int t=maxsubmin(i,to+1); if(t<=s) to+=1; else{ maxto[i]=to; break; } } } plant_tree(1,1,n+1);//线段树dp，建树 repin(i,1,n){//dp，用线段树优化 if(n-i+1<l) continue; int now=query(1,i,i); if(now==INT) continue; int to=i+l-1; if(maxto[i]<to) continue; update_leaf(1,to+1,maxto[i]+1,now+1); } int ans=query(1,n+1,n+1); if(ans==INT) ans=-1; printf("%d\n",ans); }