codeforces 487B Strip

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解决一个关于将含有数值的纸条进行最优分割的问题，确保每段至少包含l个数且最大最小值之差不超过s。使用单调队列优化计算过程。

http://www.elijahqi.win/archives/1355
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.

Examples
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.

强啊这我哪会啊

膜了一番 zhx的题解发现他单调队列用的real娴熟首先可以利用单调队列求出L【i】表示以这个数为右端点他的左端点可以延伸到哪里 dp的时候我用单调队列维护一下每次做的时候先把i-l这个位置加入并且把末端弹出因为要求最小长度是l 然后再用队首的值去更新我的答案有可能无解就保持原数组inf就可以输出的时候判断一下就行

#include<cstdio>
#include<deque>
#include<cstring>
#define inf 0x7f7f7f7f
#define N 110000
using namespace std;
inline char gc(){
    static char now[1<<16],*S,*T;
    if(S==T){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF; }
    return *S++;
}
inline int read(){
    int x=0,f=1;char ch=gc();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=gc();}
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x*f;
}
struct node{
    int id,value;
};
deque<node> q,q1,q2;
int a[N],L[N],l,s,n,f[N];
int main(){
//  freopen("cf.in","r",stdin);
    n=read();s=read();l=read();
    for (int i=1;i<=n;++i) a[i]=read();int p=0;
    for (int i=1;i<=n;++i){
        while (!q1.empty()&&a[i]+s<q1.front().value) p=q1.front().id+1,q1.pop_front();
        while (!q2.empty()&&a[i]-s>q2.front().value) p=q2.front().id+1,q2.pop_front();
        L[i]=p;node tmp;tmp.id=i;tmp.value=a[i];
        while (!q1.empty()&&a[i]>q1.back().value) q1.pop_back();q1.push_back(tmp);
        while (!q2.empty()&&a[i]<q2.back().value) q2.pop_back();q2.push_back(tmp);
    }memset(f,0x7f,sizeof(f));f[0]=0;
    for (int i=l;i<=n;++i){
        while (!q.empty()&&q.back().value>f[i-l]) q.pop_back();
        node tmp;tmp.value=f[i-l];tmp.id=i-l;q.push_back(tmp);
        while (!q.empty()&&q.front().id<L[i]-1) q.pop_front();
        if (!q.empty()&&q.front().value!=inf) f[i]=q.front().value+1;
    }
    if (f[n]==inf) printf("-1");else printf("%d\n",f[n]);
    return 0;
}