Codeforces 13C. Sequence scrolling array + discretization

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C. Sequencetime limit per test1 secondmemory limit per test64 megabytesinputstandard inputoutputstandard output

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 was allowed to increase the value of any number By 

The Sequence  a  is called non-decreasing if  a 1 ? ≤? a 2 ? ≤?...? ≤? a N  holds, Where 

Input

The first line of the input contains a 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 9 by absolute value.

Output

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

Sample Test (s) input

53 2-1 2 11

Output

4

Input

52 1 1) 1 1

Output

1

Give you n number, each operation can make a number plus 1 or minus 1, ask at least how many operations make this n number is non-decrement. Non-descending definition:

a 1 ? ≤? a 2 ? ≤?...? ≤? a N

The minimum number of requests must be made up of the original number.

DP[I][J] Represents the number of previous I to a[j] as the end of the minimum operation to meet the requirements, but the maximum number of topics given is 10^9, two-dimensional array of J elements can not open so large, so need to be discretized, changed to the number of the first I with the number of J as the end of the minimum operation to meet the requirements.

Dp[i][j]=min (Dp[i][j], the smallest operation of the first J number of the i-1 position +fabs (B[j]-a[i]) b array is the original input of the array of a well-ordered and discretized

Since B arrays are ordered from small to large, the number of J in the first position must be greater than the number of i-1 in the first position.

Finally dp[5000][5000] can not be saved, then this two-dimensional array will have to roll.

780 MS 0 kb#include<stdio.h> #include <math.h> #include <string.h> #include <algorithm># Define ll __int64using namespace Std;const ll inf = 1ll<<62;//inf to take great 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;//discretization 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]);//Take the minimum in a tree with a different height from the previous position             Dp[i&1][j]=min (Dp[i&1][j],minn+fabs (B[j]-a[i]));//The current position equals the first J tree in the previous position spend minimum + this position is the cost of the J tree} 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]);//Take the last position is the first J tree is the minimum cost of printf ("%i64d\n", ans); } return 0;}

Codeforces 13C. Sequence scrolling array + discretization