【dp】51nod 1270 数组的最大代价

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
//#pragma comment(linker, "/STACK:102400000,102400000")

const double PI = acos(-1.0);
const double eps = 1e-6;
const int INF=0x3f3f3f3f;
const LL mod = 1e9+7;
const int N = 50000+10;
const int M = 2500000;

int a
;
int dp
[2];
int main()
{
int n;
scanf("%d",&n);
for(int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
if(i == 0)
dp[i][0] = dp[i][1] = 0;
else
{
dp[i][0] = max(dp[i-1][0],dp[i-1][1]+a[i-1]-1);
dp[i][1] = max(dp[i-1][0]+a[i]-1,dp[i-1][1]+abs(a[i]-a[i-1]));
}
}
printf("%d\n",max(dp[n-1][0],dp[n-1][1]));
return 0;
}