[题目链接] https://www.luogu.org/problemnew/show/P4331
给定一个整数序列\(a_1, a_2, ··· , a_n,\)求出一个递增序列\(b_1 < b_2 < ··· < b_n\),使得序列\(a_i\)和\(b_i\)的各项之差的绝对值之和|\(a_1 - b_1| + |a_2 - b_2| + ··· + |a_n - b_n|\)最小。?
[题解] https://www.cnblogs.com/HNYLMSTea/p/10386117.html
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF=1e9+7;
inline LL read(){
register LL x=0,f=1;register char c=getchar();
while(c<48||c>57){if(c==‘-‘)f=-1;c=getchar();}
while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
return f*x;
}
const int MAXN=1e6+5;
int val[MAXN],rt[MAXN],size[MAXN],tot[MAXN],ls[MAXN],rs[MAXN],dis[MAXN],l[MAXN],r[MAXN];
int n,now;LL ans;
inline int Merge(int x,int y){
if(!x||!y) return x+y;
if(val[x]<val[y]) swap(x,y);
rs[x]=Merge(rs[x],y);
size[x]=size[ls[x]]+size[rs[x]]+1;
if(dis[rs[x]]>dis[ls[x]]) swap(ls[x],rs[x]);
dis[x]=dis[rs[x]]+1;
return x;
}
int main(){
n=read();
dis[0]=-1;
for(int i=1;i<=n;i++){
val[i]=read()-i;
now++;
l[now]=r[now]=rt[now]=i;
tot[now]=size[rt[now]]=1;
while(now>1&&val[rt[now-1]]>val[rt[now]]){
--now;
tot[now]+=tot[now+1];
r[now]=r[now+1];
rt[now]=Merge(rt[now],rt[now+1]);
while(size[rt[now]]*2>(tot[now]+1))
rt[now]=Merge(ls[rt[now]],rs[rt[now]]);
}
}
for(int i=1;i<=now;i++)
for(int j=l[i];j<=r[i];j++)
ans+=1ll*abs(val[j]-val[rt[i]]);
printf("%lld\n",ans);
for(int i=1;i<=now;i++)
for(int j=l[i];j<=r[i];j++)
printf("%d ",val[rt[i]]+j);
}
P4331 [BOI2004]Sequence 数字序列 (左偏树)
原文:https://www.cnblogs.com/lizehon/p/10392267.html