https://github.com/OI-wiki/libs/blob/master/集训队历年论文/
国家集训队2005论文集
黄源河--左偏树的特点及其应用
LuoguP4331 [BOI2004]Sequence 数字序列
分析性质后发现需要一个能删,能合的数据结构,左偏树即可,用堆维护
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int a = (b); (a) <= (c); ++(a))
#define nR(a,b,c) for(register int a = (b); (a) >= (c); --(a))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Swap(a,b) ((a) ^= (b) ^= (a) ^= (b))
#define ll long long
#define u32 unsigned int
#define u64 unsigned long long
#define ON_DEBUGG
#ifdef ON_DEBUGG
#define D_e_Line printf("
----------
")
#define D_e(x) cout << (#x) << " : " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt", "r", stdin)
#define FileSave() freopen("out.txt", "w", stdout)
#include <ctime>
#define TIME() fprintf(stderr, "
time: %.3fms
", clock() * 1000.0 / CLOCKS_PER_SEC)
#else
#define D_e_Line ;
#define D_e(x) ;
#define Pause() ;
#define FileOpen() ;
#define FileSave() ;
#define TIME() ;
//char buf[1 << 21], *p1 = buf, *p2 = buf;
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
#endif
using namespace std;
struct ios{
template<typename ATP>inline ios& operator >> (ATP &x){
x = 0; int f = 1; char ch;
for(ch = getchar(); ch < '0' || ch > '9'; ch = getchar()) if(ch == '-') f = -1;
while(ch >= '0' && ch <= '9') x = x * 10 + (ch ^ '0'), ch = getchar();
x *= f;
return *this;
}
}io;
template<typename ATP>inline ATP Max(ATP a, ATP b){
return a > b ? a : b;
}
template<typename ATP>inline ATP Min(ATP a, ATP b){
return a < b ? a : b;
}
template<typename ATP>inline ATP Abs(ATP a){
return a < 0 ? -a : a;
}
const int N = 1e6 + 7;
struct Node {
int rt, l, r, siz;
long long val;
} sta[N];
int top;
int ch[N][2], dis[N];
int a[N];
#define ls ch[x][0]
#define rs ch[x][1]
inline int Merge(int x, int y) {
if(!x || !y) return x | y;
if(a[x] < a[y]) Swap(x, y);
rs = Merge(rs, y);
if(dis[rs] > dis[ls]) Swap(ls, rs);
dis[x] = dis[rs] + 1;
return x;
}
inline int Del(int x) {
return Merge(ls, rs);
}
int main() {
int n;
io >> n;
dis[0] = -1;
R(i,1,n){
io >> a[i];
a[i] -= i;
}
R(i,1,n){
sta[++top] = (Node){ i, i, i, 1, a[i]};
while(top != 1 && sta[top - 1].val > sta[top].val){
--top;
sta[top].rt = Merge(sta[top].rt, sta[top + 1].rt);
sta[top].siz += sta[top + 1].siz;
sta[top].r = sta[top + 1].r;
while(sta[top].siz > (sta[top].r - sta[top].l + 2) / 2){
--sta[top].siz;
sta[top].rt = Del(sta[top].rt);
}
sta[top].val = a[sta[top].rt];
}
}
int tot = 1;
long long ans = 0;
R(i,1,n){
if(sta[tot].r < i) ++tot;
ans += Abs(sta[tot].val - a[i]);
}
printf("%lld
", ans);
tot = 1;
R(i,1,n){
if(sta[tot].r < i) ++tot;
printf("%lld ", sta[tot].val + i);
}
return 0;
}
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