BZOJ 1742 [Usaco2005 nov]Grazing on the Run 边跑边吃草【区间DP】

这道题就当做复习区间DP的板子题吧QvQ

$f[i][j][0/1]$表示当前已经吃完了区间 $[i,j]$内的草并且当前停留在最左边与最右边

方程随便推一下就出来了QvQ：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define db double
#define sg string
#define ll long long
#define rep(i,x,y) for(ll i=(x);i<=(y);i++)
#define red(i,x,y) for(ll i=(x);i>=(y);i--)
using namespace std;

const ll N=1e3+5;
const ll Inf=1e18;

ll n,m,p[N],f[N][N][2];

inline ll read() {
    ll x=0;char ch=getchar();bool f=0;
    while(ch>'9'||ch<'0'){if(ch=='-')f=1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return f?-x:x;
}

int main() {
	n=read(),m=read();
	
	rep(i,1,n) p[i]=read();
	
	p[++n]=m;
	
	sort(p+1,p+1+n);
	
	rep(i,1,n) f[i][i][0]=f[i][i][1]=abs(m-p[i])*n;
	
	rep(len,2,n) rep(i,1,n+1-len) {
		ll j=i+len-1;
		f[i][j][0]=min(f[i+1][j][0]+(p[i+1]-p[i])*(n-j+i),f[i+1][j][1]+(p[j]-p[i])*(n-j+i));
		f[i][j][1]=min(f[i][j-1][0]+(p[j]-p[i])*(n-j+i),f[i][j-1][1]+(p[j]-p[j-1])*(n-j+i));
	}
	
	printf("%lld\n",min(f[1][n][0],f[1][n][1]));

	return 0;
}