CodeForce 990 E. Post Lamps

最新推荐文章于 2022-05-26 19:50:53 发布

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本文介绍了一种路灯布局优化算法，旨在解决如何在街道上有效布置不同功率的路灯以覆盖整个街区的问题。考虑到某些位置可能不可用，算法需找出最低成本方案来确保整个街区被照亮。

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Adilbek's house is located on a street which can be represented as the OX axis. This street is really dark, so Adilbek wants to install some post lamps to illuminate it. Street has $n$ positions to install lamps, they correspond to the integer numbers from $0$ to $n - 1$ on the OX axis. However, some positions are blocked and no post lamp can be placed there.

There are post lamps of different types which differ only by their power. When placed in position $x$ , post lamp of power $l$ illuminates the segment $[x; x + l]$ . The power of each post lamp is always a positive integer number.

The post lamp shop provides an infinite amount of lamps of each type from power $1$ to power $k$ . Though each customer is only allowed to order post lamps of exactly one type. Post lamps of power $l$ cost $a_{l}$ each.

What is the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street? If some lamps illuminate any other segment of the street, Adilbek does not care, so, for example, he may place a lamp of power $3$ in position $n - 1$ (even though its illumination zone doesn't completely belong to segment $[0; n]$ ).

Input

The first line contains three integer numbers $n$ , $m$ and $k$ ( $1 \leq k \leq n \leq 10^{6}$ , $0 \leq m \leq n$ ) — the length of the segment of the street Adilbek wants to illuminate, the number of the blocked positions and the maximum power of the post lamp available.

The second line contains $m$ integer numbers $s_{1}, s_{2}, \dots, s_{m}$ ( $0 \leq s_{1} < s_{2} < \dots s_{m} < n$ ) — the blocked positions.

The third line contains $k$ integer numbers $a_{1}, a_{2}, \dots, a_{k}$ ( $1 \leq a_{i} \leq 10^{6}$ ) — the costs of the post lamps.

Output

Print the minimal total cost of the post lamps of exactly one type Adilbek can buy to illuminate the entire segment $[0; n]$ of the street.

If illumintaing the entire segment $[0; n]$ is impossible, print -1.

Examples
inputCopy
6 2 3
1 3
1 2 3
outputCopy
6
inputCopy
4 3 4
1 2 3
1 10 100 1000
outputCopy
1000
inputCopy
5 1 5
0
3 3 3 3 3
outputCopy
-1
inputCopy
7 4 3
2 4 5 6
3 14 15
outputCopy
-1

题解：
n/1+n/2+n/3+...+n/n=n*logn
直接暴力即可。
代码：
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=1e6+7;
bool vis[maxn];
int n,m,k,last[maxn];
int main()
{
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=m;i++)
    {
        int x;scanf("%d",&x);
        vis[x]=1;
    }
    if(vis[0])
    {
        printf("-1\n");
        return 0;
    }
    int la=0;
    for(int i=1;i<=n;i++)//last保存要想照亮该点所能安装等的最大坐标
    {
        if(!vis[i])la=i;
        last[i]=la;
    }
    ll ans=-1;
    for(int i=1;i<=k;i++)
    {
        int x;scanf("%d",&x);
        ll sum=0;
        for(int j=0;j<n; )
        {
            sum+=x;
            int id;
            if(j+i>n)id=n;
            else id=last[i+j];
            if(id<=j)
            {
                sum=-1;break;
            }
            j=id;
        }
        if(sum!=-1)
        {
            if(ans==-1)ans=sum;
            else ans=min(ans,sum);
        }
    }
    printf("%lld\n",ans);
    return 0;
}