poj3709

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题解

容易得到的dp方程

  • 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>=x;--i)
#define ll long long
using namespace std;
const int N=1e6+7;
ll dp[N],a[N],sum[N];
int n,m,q[N];
inline void init(){
	rep(i,1,n){
		scanf("%lld",&a[i]);
		sum[i]=sum[i-1]+a[i];
	}
}
inline ll Y(int x,int y){return dp[x]+a[x+1]*x-sum[x]-(dp[y]+a[y+1]*y-sum[y]);}
inline ll X(int x,int y){return a[x+1]-a[y+1];}
int main(){
	scanf("%d",&n);
	while(~scanf("%d%d",&n,&m)){
		init();
		int head=1,tail=1;
		rep(i,1,2*m-1)dp[i]=sum[i]-a[1]*i;
		rep(i,2*m,n){                   //确保第一段一定是1开始的且长度为m
			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("%lld\n",dp[n]);
	}
	return 0;
}