hdu3480

发表于 | 阅读数 次

链接:hdu3480

题解

  • 数组排个序,容易得到dp[j][i]表示前i个数分成j组的最小值

  • 显然可以斜率优化

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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e4+7;
const int M=5e3+7;
int dp[M][N],a[N];
int n,m,T,q[N];
inline int Pow(ll x){return x*x;}
inline void init(){
	scanf("%d%d",&n,&m);
	rep(i,1,n)scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	rep(i,1,n)dp[1][i]=Pow(a[i]-a[1]);
}
inline int Y(int x,int y,int c){return dp[c][x]+Pow(a[x+1])-(dp[c][y]+Pow(a[y+1]));}
inline int X(int x,int y){return a[x+1]-a[y+1]<<1;}
int main(){
	scanf("%d",&T);
	rep(cnt,1,T){
		init();
		rep(j,2,m){
			int head=1,tail=1;
			rep(i,2,n){
				while(head<tail&&Y(q[tail],i-1,j-1)*X(q[tail-1],q[tail])<=Y(q[tail-1],q[tail],j-1)*X(q[tail],i-1))--tail;
				q[++tail]=i-1;
				while(head<tail&&Y(q[head+1],q[head],j-1)<=X(q[head+1],q[head])*a[i])++head;
				dp[j][i]=dp[j-1][q[head]]+Pow(a[i]-a[q[head]+1]);
			}
		}
		printf("Case %d: %d\n",cnt,dp[m][n]);
	}
	return 0;
}