https://codeforces.com/gym/101473/attachments

题意

环形DP

\(dp[i]\)表示\(1～i\)被处理完最少需要覆盖多长的补丁

用二分找出上一个转移位置

然后枚举序列情况即可

复杂度\(O(n^2log n)\)

#include<cstdio>
#include<iostream>
#include<algorithm>
#define INF 2000000007
#define N 2005
using namespace std;
int n,c,t1,t2,a[N],d[N];
int dp[N];
int main()
{
    scanf("%d%d%d%d",&n,&c,&t1,&t2);
    for (int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    sort(a+1,a+n+1);
    int ans=INF;
    for (int t=1;t<=n;t++)
    {
        for (int i=1;i<=n;i++)
            d[i]=((a[i]-a[t])%c+c)%c;
        sort(d+1,d+n+1);
        dp[0]=0;
        for (int i=1;i<=n;i++)
        {
            dp[i]=dp[lower_bound(d+1,d+n+1,d[i]-t1)-d-1]+t1;
            dp[i]=min(dp[i],dp[lower_bound(d+1,d+n+1,d[i]-t2)-d-1]+t2);
        }
        ans=min(ans,dp[n]);
    }
    cout << ans << endl;
    return 0;
}

Gym 101473I Patches

原文：https://www.cnblogs.com/GK0328/p/13373646.html