BC#86.1002(HDOJ5805)NanoApe Loves Sequence

记录一个菜逼的成长。。

求出前i个数里相邻差值的最大值fi，i到n里相邻差值的最大值gi​​，那么ans=∑i=1nmax(∣Ai−1−Ai+1∣,fi−1,gi+1)。
时间复杂度O(n)。
一道水题。。麻痹脑子抽了写了好久。。

#include <cstdio>#include <iostream>#include <cstring>#include <string>#include <algorithm>#include <cstdlib>#include <vector>#include <set>#include <map>#include <queue>#include <list>#include <deque>#include <cctype>#include <bitset>#include <cmath>using namespace std;#define cl(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> PII;typedef pair<LL,LL> PLL;const int INF = 0x3f3f3f3f;const int maxn = 100000 + 10;//lmx[i] 表示1-i相邻差的最大值，rmx[i]表示i-n相邻差的最大值。int a[maxn],lmx[maxn],rmx[maxn];int main(){    int T;scanf("%d",&T);    while(T--){        int n;        scanf("%d",&n);        lmx[1] = rmx[n] = 0;        for( int i = 1; i <= n; i++ ){            scanf("%d",a+i);            if(i > 1){                lmx[i] = max(lmx[i-1],abs(a[i] - a[i-1]));            }        }        for( int i = n - 1; i > 0; i-- ){            rmx[i] = max(rmx[i+1], abs(a[i+1] - a[i]));        }        LL ans = (LL)(lmx[n-1] + rmx[2]);//删去最左和最右两个数独立讨论        for( int i = 2; i < n; i++ ){            ans += max(abs(a[i+1] - a[i-1]),max(lmx[i-1],rmx[i+1]));        }        printf("%lld\n",ans);    }    return 0;}