ARC 119 补题记录
作者：互联网
这把感觉质量很高。
\(E\)
\(E\)比较简单所以先写个\(E\),考虑就一个置换操作来说改变的只有两端的值。
考虑\(|a_i - a_{i - 1}|\)变成区间，则我们考虑分类讨论，发现只有当\(a_{i + 1} > a_{i}\)且\(a_r > a_{r + 1}\)还有\(a_{i + 1} < a_{i}\)且\(a_r < a_{r + 1}\)时，交换操作会带来一些贡献，这个贡献是两倍交集。两种情况可以反转序列来做。、（注意单独考虑\(1\)和\(n\)）的情况。
E
#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
#define N 400005
ll n,a[N],sum,ans;
struct P{ll l,r;}b[N];
inline bool operator < (P a,P b){return a.l < b.l;}
inline ll abs(ll a){return a >= 0? a : -a;}
int main(){
scanf("%lld",&n);
for(int i = 1;i <= n;++i)
scanf("%lld",&a[i]);
// for(int i = 1;i <= n;++i)
// std::cout<<a[i]<<" ";
for(int i = 2;i <= n;++i)
sum += abs(a[i] - a[i - 1]);
ans = sum;
// std::cout<<sum<<std::endl;
for(int i = 2;i <= n - 1;++i)
ans = std::min(ans,sum - abs(a[i + 1] - a[i]) + abs(a[i + 1] - a[1]));
for(int i = 2;i <= n - 1;++i)
ans = std::min(ans,sum - abs(a[i - 1] - a[i]) + abs(a[i - 1] - a[n]));
//(al,al + 1) (ar,ar + 1)
ll cnt = 0;
for(int i = 1;i <= n - 1;++i)
if(a[i] < a[i + 1])
b[++cnt].l = a[i],b[cnt].r = a[i + 1];
std::sort(b + 1,b + cnt + 1);
ll maxr = b[1].r;
for(int i = 2;i <= cnt;++i){
// std::cout<<b[i].l<<" "<<b[i].r<<std::endl;
ans = std::min(ans,sum - 2 * (std::min(maxr,b[i].r) - b[i].l));
maxr = std::max(b[i].r,maxr);
}
std::reverse(a + 1, a + n + 1);
cnt = 0;
for(int i = 1;i <= n - 1;++i)
if(a[i] < a[i + 1])
b[++cnt].l = a[i],b[cnt].r = a[i + 1];
std::sort(b + 1,b + cnt + 1);
maxr = b[1].r;
for(int i = 2;i <= cnt;++i){
ans = std::min(ans,sum - 2 * (std::min(maxr,b[i].r) - b[i].l));
maxr = std::max(b[i].r,maxr);
}
std::cout<<ans<<std::endl;
}
标签：return,int,ll,long,ARC,补题,inline,include,119 来源： https://www.cnblogs.com/dixiao/p/14787599.html