[JSOI2016]

• 题目
• g[i][j]表示 i->j->i 过程中i~j村庄最少消失的人数
• f[i]表示当前在i村庄，i及i之前的村庄都被治愈的总共最少消失人数

#include<bits/stdc++.h>
#define rep(i,a,b) for(ll i=(a);i<=(b);i++)
#define ll long long
using namespace std;
const ll N=3e3+10;
ll g[N][N],sum[N],f[N],a[N],n;
inline ll read(){
	ll num=0;char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))num=num*10+ch-'0',ch=getchar();
	return num;
}
int main()
{
	n=read();
	rep(i,1,n)a[i]=read();
	rep(i,1,n)sum[i]=a[i]+sum[i-1];
	rep(len,1,n-1){
		rep(i,1,n-len){
			ll j=i+len;
			g[i][j]=g[i+1][j]+min((sum[j]-sum[i])*2,3*a[i]*(j-i)+sum[j]-sum[i]);
		}
	}
	memset(f,0x3f,sizeof(f));f[0]=0;
	rep(i,1,n){
		rep(j,0,i-1){
			f[i]=min(f[i],f[j]+g[j+1][i]+(sum[n]-sum[i])*(3*(i-j-1)+i-j+1));
		}
	}
	cout<<f[n];
	return 0;
}