Tenka1 Programmer Contest C

链接 Tenka1 Programmer Contest C - Align

给定一个序列，要求重新排列最大化\(\sum_{i=2}^{i=n} |a_i-a_{i-1}|\)，\(n\leq 10^5\)
小清新贪心，首先把最大的先放好，然后依次考虑下面四种决策：
左边放最小，右边放最小，左边放最大，右边放最大。
每次取\(max\)并更新左右端点，这样一定能取到最大最小的波浪形态，最大值旁边放两个最小不会更差。

#include<bits/stdc++.h>
#define R register int
#define ll long long
using namespace std;
const int N=100001;
int n,w[N],res[N],le,ri,z,y;ll ans;
int gi(){R x=0,k=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')k=-1,c=getchar();while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+c-'0',c=getchar(); return x*k;
}
int main(){n=gi();for(R i=1;i<=n;++i)w[i]=gi();sort(w+1,w+n+1);le=2,ri=n,z=w[1],y=w[1];while(le<=ri){R p=abs(w[le]-z),q=abs(w[le]-y),a=abs(w[ri]-z),b=abs(w[ri]-y),Mx=max(max(p,q),max(a,b));if(Mx==p)z=w[le],le++;else if(Mx==q)y=w[le],le++;else if(Mx==a)z=w[ri],ri--;else y=w[ri],ri--;ans+=Mx;}cout<<ans<<endl;return 0;
}

转载于:https://www.cnblogs.com/Tyher/p/9863504.html

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Tenka1 Programmer Contest C - Align

链接 Tenka1 Programmer Contest C - Align

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