论文补全计划

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("\n----------\n")
#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, "\ntime: %.3fms\n", 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\n", ans);
	tot = 1;
	R(i,1,n){
		if(sta[tot].r < i) ++tot;
		printf("%lld ", sta[tot].val + i);
	}
	return 0;
}

咕咕咕

posted @ 2019-10-11 10:58  邱涵的秘密基地  阅读(158)  评论(0编辑  收藏  举报