BestCoder Round #86 1002 &HDU 5805 ——NanoApe Loves Sequence

雨点打透心脏的1/2处 2022-07-19 01:12 101阅读 0赞

### 题意 ###

给定一个数列，随机从该数列里删除一个数，求该数列的的相邻之间的绝对值的最大值的和。

### 思路 ###

两个数列来分别维护i位前面的相邻的绝对值的最大以及i后面的最大，枚举每一个可能删除的数，然后分别从左右和要删除的这个空档里娶一个最大的值。

code：

#include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <algorithm>
    #include <cstring>
    #include <sstream>
    #include <string>
    #include <vector>
    #include <list>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #include <bitset>
    
    using namespace std;
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef long double ld;
    
    const int INF=0x3fffffff;
    const int inf=-INF;
    const int N=100005;
    const int M=2005;
    const int mod=1000000007;
    const double pi=acos(-1.0);
    
    #define cls(x,c) memset(x,c,sizeof(x))
    #define cpy(x,a) memcpy(x,a,sizeof(a))
    #define ft(i,s,n) for (int i=s;i<=n;i++)
    #define frt(i,s,n) for (int i=s;i>=n;i--)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define lrt rt<<1
    #define rrt rt<<1|1
    #define middle int m=(r+l)>>1
    #define lowbit(x) (x&-x)
    #define pii pair<int,int>
    #define mk make_pair
    #define IN freopen("in.txt","r",stdin)
    #define OUT freopen("out.txt","w",stdout)
    
    int read() {
        char ch;
        while (ch = getchar(), !isdigit(ch));
        int res = ch - '0';
        while (ch = getchar(), isdigit(ch))
            res = res * 10 + ch - '0';
        return res;
    }
    //++++++++++++密++++++++++++++封++++++++++++++++++++线
    
    ll v[N],s1[N],s2[N];
    int main()
    {
        int T=read();
        ft(ca,1,T){
            int n=read();
            memset(s1,0,sizeof(s1));
            memset(s2,0,sizeof(s2));
            ft(i,1,n){
                scanf("%I64d",v+i);
                if (i>1){
                    ll tq=abs(v[i]-v[i-1]);
                    s1[i]=max(s1[i-1],tq);
                }
            }
            frt(i,n,1){
                if (i<n){
                    ll tq=abs(v[i]-v[i+1]);
                    s2[i]=max(s2[i+1],tq);
                }
            }
            ll ans=s2[2]+s1[n-1];
            ft(i,2,n-1){
                ll tq=abs(v[i+1]-v[i-1]);
                tq=max(tq,s1[i-1]);
                tq=max(tq,s2[i+1]);
                ans+=tq;
            }
            printf("%I64d\n",ans);
        }
    }