Description
Input
Output
Solution
比较明显的DP
设(f_{i,j})表示做到第i行,分了j个矩形的方案数
转移明显:
(f_{i,j}=max(f_{k,j-1}+(i-k)*(r_{i}-l_{i})))
暴力做的话是(O(n^{2}k))的,不能接受
考虑斜率优化,就能过了
时间复杂度(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;
}