Codeforces Round #276 (Div. 1) D. Kindergarten dp

D. Kindergarten

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/484/D

Description

In a kindergarten, the children are being divided into groups. The teacher put the children in a line and associated each child with his or her integer charisma value. Each child should go to exactly one group. Each group should be a nonempty segment of consecutive children of a line. A group's sociability is the maximum difference of charisma of two children in the group (in particular, if the group consists of one child, its sociability equals a zero).

The teacher wants to divide the children into some number of groups in such way that the total sociability of the groups is maximum. Help him find this value.

Input

The first line contains integer n — the number of children in the line (1 ≤ n ≤ 106).

The second line contains n integers ai — the charisma of the i-th child ( - 109 ≤ ai ≤ 109).

Output

Print the maximum possible total sociability of all groups.

Sample Input

5
1 2 3 1 2

Sample Output

3

HINT

 

题意

给你n个数，然后让你分成若干组，只有连续的才能分成一组

然后每组的分数是，这组的最大值减去这组的最小值

问你最大能拿多少分

题解：

很显然单调的必然会分在一个组，这儿有一个问题就是拐点怎么办，分在左边还是右边？

那就dp咯

代码

 

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 1000005
#define mod 10007
#define eps 1e-5
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

ll dp1[maxn];
ll dp2[maxn];
ll dp3[maxn];
int a[maxn];
int main()
{
    int n=read();
    for(int i=0;i<=n;i++)
        dp1[i]=dp2[i]=dp3[i]=-infll;
    for(int i=1;i<=n;i++)
        a[i]=read();
    dp1[1]=a[1];
    dp2[1]=-a[1];
    dp3[1]=0;
    for(int i=2;i<=n;i++)
    {
        dp3[i]=max(dp3[i-1],max(dp1[i-1]-a[i],dp2[i-1]+a[i]));
        dp1[i]=max(dp1[i-1],dp3[i-1]+a[i]);
        dp2[i]=max(dp2[i-1],dp3[i-1]-a[i]);
    }
    cout<<dp3[n]<<endl;
    return 0;
}