LOJ6435「PKUSC2018」星际穿越（倍增dp，主席树）

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我是靠谱客的博主悦耳小刺猬，这篇文章主要介绍LOJ6435「PKUSC2018」星际穿越（倍增dp，主席树），现在分享给大家，希望可以做个参考。

传送门：https://loj.ac/problem/6435

这里讲的是倍增写法，另有主席树解法

可以贪心的发现最优的走法必然是从 $x$ 一直往左跳或者先向右跳一步再一直往左跳
一个暴力：
$f_{i,j}$ 表示从 $[i, n]$ 这些点出发，向左跳 $j$ 步能转移到的最左端点
$O(n^2)$

很显然这个式子可以用倍增优化， $f_{i,j}$ 表示从 $[i, n]$ 出发走 $2^j$ 步能到达的最左点
问题在于如何计算答案
另记 $g_{i,j}$ 表示从 $[i, n]$ 这些点出发 $走2^j$ 步后，到 $f_{i,j-1},i-1]$ 这些点的最短路之和

简单推导可以得到
$g_{i,j} = g_{i,j-1}+g_{(f_{i,j-1}),j-1} + (f_{i,j-1} - f_{i,j})*2^{(j-1)}$
转移的时候直接用即可

注意边界

复制代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair<int,int>
#define Pil pair<int,ll>
#define Pli pair<ll,int>
#define Pll pair<ll,ll>
#define pb push_back 
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
int rd()
{
    int sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48,c = getchar();
    if(flag) return sum;
    else return -sum;
}
int n,L[301000],Q,mn[301000][20];
int f[301000][20];
ll sigma[301000][20];
int min(int a,int b){return a<b?a:b;}
int max(int a,int b){return a>b?a:b;}
int ask(int l,int r)
{
	int k = log2(r-l+1);
	return min(mn[l][k],mn[r-(1<<k)+1][k]);
}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void init()
{
	n = rd();
	rep(i,2,n) L[i] = rd();
	rep(i,1,n) mn[i][0] = L[i];
	for(int j = 1;(1<<j) <= n;++j)
		for(int i = 1;i + (1<<j) - 1 <= n;++i)
			mn[i][j] = min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
    rep(i,1,n) f[i][0] = ask(i,n),sigma[i][0] = i-f[i][0];
    for(int j = 1;(1<<j) <= n;++j)
    	for(int i = 1;i <= n;++i)
    	    if(f[i][j-1])
    		{
    			f[i][j] = f[f[i][j-1]][j-1];
	    		sigma[i][j] = sigma[i][j-1] + sigma[f[i][j-1]][j-1]
    			     + 1ll*(f[i][j-1]-f[i][j])*(1<<(j-1));
    		}
}
ll solve(int l,int r)
{
	if(L[r] <= l) return r-l;
	ll sum1 = r-L[r];
	int pl = L[r],sum2 = 2;
	for(int j = 19;j>=0;j--)
		if(f[pl][j] > l)
		{
			sum1 += (sigma[pl][j]+1ll*(pl-f[pl][j])*(sum2-1));
			pl = f[pl][j];
			sum2 += 1<<j;
		}
	sum1 += 1ll*(pl-l)*sum2;
	return sum1;
}
void work()
{
	Q = rd();
	rep(i,1,Q)
	{
		int l = rd(),r = rd(),x = rd();
		ll ans = solve(l,x) - solve(r+1,x),tmp = gcd(ans,r-l+1);
		printf("%lld/%lldn",ans/tmp,1ll*(r-l+1)/tmp);
	}
}
int main()
{
	init();
	work();
    return 0;
}

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#include<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair<int,int>
#define Pil pair<int,ll>
#define Pli pair<ll,int>
#define Pll pair<ll,ll>
#define pb push_back 
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
int rd()
{
    int sum = 0;char c = getchar();bool flag = true;
    while(c < '0' || c > '9') {if(c == '-') flag = false;c = getchar();}
    while(c >= '0' && c <= '9') sum = sum * 10 + c - 48,c = getchar();
    if(flag) return sum;
    else return -sum;
}
int n,L[301000],Q,mn[301000][20];
int f[301000][20];
ll sigma[301000][20];
int min(int a,int b){return a<b?a:b;}
int max(int a,int b){return a>b?a:b;}
int ask(int l,int r)
{
	int k = log2(r-l+1);
	return min(mn[l][k],mn[r-(1<<k)+1][k]);
}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void init()
{
	n = rd();
	rep(i,2,n) L[i] = rd();
	rep(i,1,n) mn[i][0] = L[i];
	for(int j = 1;(1<<j) <= n;++j)
		for(int i = 1;i + (1<<j) - 1 <= n;++i)
			mn[i][j] = min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);
    rep(i,1,n) f[i][0] = ask(i,n),sigma[i][0] = i-f[i][0];
    for(int j = 1;(1<<j) <= n;++j)
    	for(int i = 1;i <= n;++i)
    	    if(f[i][j-1])
    		{
    			f[i][j] = f[f[i][j-1]][j-1];
	    		sigma[i][j] = sigma[i][j-1] + sigma[f[i][j-1]][j-1]
    			     + 1ll*(f[i][j-1]-f[i][j])*(1<<(j-1));
    		}
}
ll solve(int l,int r)
{
	if(L[r] <= l) return r-l;
	ll sum1 = r-L[r];
	int pl = L[r],sum2 = 2;
	for(int j = 19;j>=0;j--)
		if(f[pl][j] > l)
		{
			sum1 += (sigma[pl][j]+1ll*(pl-f[pl][j])*(sum2-1));
			pl = f[pl][j];
			sum2 += 1<<j;
		}
	sum1 += 1ll*(pl-l)*sum2;
	return sum1;
}
void work()
{
	Q = rd();
	rep(i,1,Q)
	{
		int l = rd(),r = rd(),x = rd();
		ll ans = solve(l,x) - solve(r+1,x),tmp = gcd(ans,r-l+1);
		printf("%lld/%lldn",ans/tmp,1ll*(r-l+1)/tmp);
	}
}
int main()
{
	init();
	work();
    return 0;
}