HDU 5805/BC 86B NanoApe Loves Sequence

题解：首先计算出每两个相邻数之间的差值，统计出其中最大的三个。然后枚举删每一个点，对于删除当前点，首先删除当前点与左右两个点之间的距离 (只有删除的距离为最大的两个时才会有影响)，然后处理没被删掉的最大值与新生成的左右两个点之间的差值作比较，取最大值加到答案上。

对于前三大的值可以开三个变量统一，为了方便也可以sort，我比赛的时候脑抽用multiset维护的 ,也就是6倍sort的时间，终判就T了...

也不明白为什么才10个样例 n才1e5 O(nlogn)的算法也能T ...又一次败在水题上..

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<deque>
#define mem(x,y) memset(x,y,sizeof(x))
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define bug puts("===========");
#define zjc puts("");
const double pi=(acos(-1.0));
const double eps=1e-8;
const ll INF=1e18+10;
const ll inf=1e9+10;
const int mod=998244353;
const int maxn=100000+5;
/*=======================================*/
namespace fastIO{
#define ll long long
inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}
//getchar->read
inline void read1(int &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(ll &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (bo)x=-x;
}
inline void read1(double &x){
char ch;int bo=0;x=0;
for (ch=getchar();ch<'0'||ch>'9';ch=getchar())if (ch=='-')bo=1;
for (;ch>='0'&&ch<='9';x=x*10+ch-'0',ch=getchar());
if (ch=='.'){
double tmp=1;
for (ch=getchar();ch>='0'&&ch<='9';tmp/=10.0,x+=tmp*(ch-'0'),ch=getchar());
}
if (bo)x=-x;
}
inline void read1(char *s){
char ch=getchar();
for (;blank(ch);ch=getchar());
for (;!blank(ch);ch=getchar())*s++=ch;
*s=0;
}
inline void read1(char &c){for (c=getchar();blank(c);c=getchar());}
//scanf->read
inline void read2(int &x){scanf("%d",&x);}
inline void read2(ll &x){
#ifdef _WIN32
scanf("%I64d",&x);
#else
#ifdef __linux
scanf("%lld",&x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
inline void read2(double &x){scanf("%lf",&x);}
inline void read2(char *s){scanf("%s",s);}
inline void read2(char &c){scanf(" %c",&c);}
inline void readln2(char *s){gets(s);}

//puts->write
inline void print1(int x){
static char buf[15];
char *p1=buf;if (!x)*p1++='0';if (x<0)putchar('-'),x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)putchar(*p1);
}
inline void println1(int x){print1(x);putchar('\n');}
inline void print1(ll x){
static char buf[25];
char *p1=buf;if (!x)*p1++='0';if (x<0)putchar('-'),x=-x;
while(x)*p1++=x%10+'0',x/=10;
while(p1--!=buf)putchar(*p1);
}
inline void print1(char c){putchar(c);}
inline void print1(char *s){while (*s)putchar(*s++);}
inline void pe(){putchar('\n');}
inline void pk(){putchar(' ');}
inline void print2(int x){printf("%d",x);}
inline void print2(char x){printf("%c",x);}
inline void print2(ll x){
#ifdef _WIN32
printf("%I64d",x);
#else
#ifdef __linux
printf("%lld",x);
#else
puts("error:can't recognize the system!");
#endif
#endif
}
#undef ll
};
using namespace fastIO;
/*==================================*/
int a[maxn];
int b[maxn];
int main()
{
int T_T;
read1(T_T);
while(T_T--){
int n;
read1(n);
for(int i=1;i<=n;i++) {
read1(a[i]);
if(i!=1) b[i-1]=abs(a[i]-a[i-1]);
}
ll ans=0;
sort(b+1,b+n);
for(int i=1;i<=n;i++){
int l=0,r=0,k=0;
if(i!=1) l=abs(a[i]-a[i-1]);
if(i!=n) r=abs(a[i]-a[i+1]);
if(i!=1&&i!=n) k=abs(a[i+1]-a[i-1]);
if(l>r) swap(l,r);
if(r==b[n-1]){
if(l==b[n-2]) ans+=max(b[n-3],k);
else ans+=max(b[n-2],k);
}else ans+=max(b[n-1],k);
}
printf("%lld\n",ans);
}
return 0;
}