Description

Input

Output

Solution

比較明顯的DP
設\(f_{i,j}\)表示做到第i行，分了j個矩形的方案數
轉移明顯：
\(f_{i,j}=max(f_{k,j-1}+(i-k)*(r_{i}-l_{i}))\)
暴力做的話是\(O(n^{2}k)\)的，不能接受
考慮斜率優化，就能過了
時間複雜度\(O(nk)\)

Code

#include <cstdio>
#include <algorithm>
#define N 20001
#define open(x) freopen(x".in","r",stdin);freopen(x".out","w",stdout);
using namespace std;
int n,k,l,r,i,j,x[N],y[N],w[N],p[N];
long long ans,f[N][110];
long double pd(int x,int y)
{
    return (f[x][i-1]-f[y][i-1])/(long double)(x-y);
}
int main()
{
    open("pyramid");
    scanf("%d%d",&n,&k);
    for (i=1;i<=n;i++)
    {
        scanf("%d%d",&x[i],&y[i]);
        w[i]=y[i]-x[i]+1;
    }
    for (i=1;i<=k;i++)
    {
        l=r=0;
        for (j=1;j<=n;j++)
        {
            while (l<r && pd(p[l],p[l+1])>w[j]) l++;
            f[j][i]=max(f[j-1][i],f[p[l]][i-1]+1ll*w[j]*(j-p[l])); 
            while (l<r && pd(p[r],j)>pd(p[r-1],p[r])) r--;
            p[++r]=j;
            ans=max(ans,f[j][i]);
        }
    }
    printf("%lld",ans);
    return 0;
}