hdu3045

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链接:hdu3045

题解

  • 显然根据题意排序后,所有集合都是连续的,a[i]表示每只cow的Moo~,sum前缀和,dp[i]表示前i只牛组成的集合的最小减少

即

  • 显然是个斜率dp
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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=4e5+7;
template <typename T>inline void read(T &x){
	char c;int sign=1;x=0;
	do{c=getchar();if(c=='-')sign=-1;}while(c<'0'||c>'9');
	do{x=x*10+c-'0';c=getchar();}while(c>='0'&&c<='9');
	x*=sign;
}
ll dp[N],sum[N],a[N];
int n,m,q[N];
inline void init(){
	rep(i,1,n)read(a[i]);
	sort(a+1,a+n+1);
	rep(i,1,n)sum[i]=a[i]+sum[i-1];
	rep(i,m,2*m-1)dp[i]=sum[i]-i*a[1];
}
inline ll Y(int x,int y){return dp[x]-sum[x]+a[x+1]*x-(dp[y]-sum[y]+a[y+1]*y);}
inline ll X(int x,int y){return a[x+1]-a[y+1];}
int main(){
	while(~scanf("%d%d",&n,&m)){
		init();
		int head=1,tail=1;
		rep(i,2*m,n){
			while(head<tail&&Y(q[tail],i-m)*X(q[tail-1],q[tail])<=Y(q[tail-1],q[tail])*X(q[tail],i-m))--tail;
			q[++tail]=i-m;
			while(head<tail&&Y(q[head+1],q[head])<=X(q[head+1],q[head])*i)++head;
			dp[i]=dp[q[head]]+sum[i]-sum[q[head]]-a[q[head]+1]*(i-q[head]);
		}
		printf("%I64d\n",dp[n]);
	}
	return 0;
}