CodeForces 13C. Sequence 滚动数组+离散化

Little Petya likes to play very much. And most of all he likes to play the following game:

He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.

The sequence a is called non-decreasing if a1?≤?a2?≤?...?≤?aN holds, where N is the length of the sequence.

Input

The first line of the input contains single integer N (1?≤?N?≤?5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 109 by absolute value.

Output

Output one integer — minimum number of steps required to achieve the goal.

Sample test(s)

input
5
3 2 -1 2 11

output
4

input
5
2 1 1 1 1

output
1

//780 ms 0 KB #include<stdio.h> #include<math.h> #include<string.h> #include<algorithm> #define ll __int64 using namespace std; const ll inf = 1ll<<62;//inf要取很大 ll dp[2][5007],a[5007],b[5007]; ll min(ll x,ll y) { return x<y?x:y; } int main() { ll n; while(scanf("%I64d",&n)!=EOF) { for(int i=1;i<=n;i++) { scanf("%I64d",&a[i]); b[i]=a[i]; } sort(b+1,b+n+1); int tot=unique(b+1,b+n+1)-b-1;//离散化 dp[1][1]=inf; for(int i=1;i<=tot;i++) { dp[1&1][i]=fabs(b[i]-a[1]); dp[2&1][i]=inf; } for(int i=2;i<=n;i++) { ll minn=inf; for(int j=1;j<=tot;j++) { minn=min(minn,dp[(i-1)&1][j]);//取上一个位置中前j颗不同高度的树中花费最小 dp[i&1][j]=min(dp[i&1][j],minn+fabs(b[j]-a[i]));//当前位置等于上一个位置中前j颗树花费最小+这个位置是第j颗树的花费 } for(int j=1;j<=tot;j++)dp[(i+1)&1][j]=inf; } ll ans=inf; for(int i=1;i<=tot;i++)ans=min(ans,dp[n&1][i]);//取最后一个位置是第j颗树是花费最小 printf("%I64d\n",ans); } return 0; }