POJ - 3666 Making the Grade

题目大意：

N个数表示每个点的海拔，题目是问你改如何花最少的花费，将N个点的海拔调整成非严格递增或非严格递减

 

分析：

dp。dp[i][j]表示前i个，j为最大数的非严格递增或非严格递减序列所需的最小花费，但是题目中的j其实是很大，
可是j的个数很少，所以dp[i][j]中j的含义改成，B中第j个元素为最大数，前i个的最少花费，状态转移方程也可以
得出了dp[i][j]=min(dp[i-1][k])+abs(a[i]-b[j])详细看代码

 

code:

#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define pb push_back
#define dbg(x) cout<<#x<<" = "<<(x)<<endl;
#define lson l,m,rt<<1
#define cmm(x) cout<<"("<<(x)<<")";
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;
const int maxn=1e5;
const int INF=0x3f3f3f3f;
const ll inf=0x7fffff;
const int mod=1e9+7;
const int MOD=10007;
//----
//define
ll a[maxn];
ll b[maxn];
ll dp[2005][2005]; 
//solve
void solve() {
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		b[i]=a[i];
	}
	sort(b+1,b+1+n);
	for(int i=1;i<=n;i++){
		ll mi=dp[i-1][1];
		for(int j=1;j<=n;j++){
			mi=min(mi,dp[i-1][j]);
			dp[i][j]=mi+llabs(a[i]-b[j]);
		}
	}
	ll mi=INF;
	for(int i=1;i<=n;i++){
		mi=min(mi,dp[n][i]);
	}
	cout<<mi<<endl;
}

int main() {
	ios_base::sync_with_stdio(0);
#ifdef debug
	freopen("in.txt","r",stdin);
//	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();
	return 0;
}