【dp】51nod 1270 数组的最大代价

只有最底端和最顶端两条路是可能最优的。

#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[N];int dp[N][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;}