P2672 [NOIP2015 Popularization Group] Salesman

greedy simulation

Portal

Let's start with this example

6
1 2 3 4 5 6
4 2 3 5 3 1

If current $X=1$ , then we can build a$F$ array
$F\left[i\right]$ means to go in and out from the entrance, and add the first$a\left[i\right]$ Home users promote sales, do not take redundant roads, and will contribute to the answer.
Use$ans$ means the current answer,
use$now$ represents the farthest point

Then $X=1$ 时: $F[i]=s[i]\ast 2+a[i]$
$F=$ { $6,6,9,13,13,13$ }
So we start fromF[i] F[i]Just pick a maximum value in$F$$[$$i$$]$
Set$now=max.id$ 即 $now=4$ (in fact, 5 or 6 is also acceptable)
$ans$ += $F[4]$
($ans$ == $13$ )
Then think about it, what will be the effect on other$What\; is\; the\; effect\; of\; F\left[i\right]$ ?
Considering$X=2$ o'clock.

young $s[i]$ < $s\left[now\right]$ is ins [ now ] s[now]left side of$s$$[$$n$$o$$w$$]$

$F[i]$ = $a[i]$

Because Keyi is going now nowSales on the road at point$n$$o$$w$
Supplement:Actually go to sales first$i$ point and go to sell$The\; fatigue\; value\; of\; point\; now$ is the same
, then$F$ becomes { 4 , 2 , 3 , / , 13 , 13 4, 2, 3, /, 13, 13 $4,2,3,/,13,13$}

young $s[i]$ > $s\left[now\right]$ is ins [ now ] s[now]Right side of$s$$[$$n$$o$$w$$]$

$F[i]=(s[i]-s[now])\ast 2+a[i]$

Because you can go two more distances forward on the original distance ( $s[i]-s[now]$) $\ast 2$ (one trip, one trip)
so$F$ becomes { 4 , 2 , 3 , / , 5 , 5 4, 2, 3, /, 5, 5 $4,2,3,/,5,5$}

Now, we start again from FFFind the ID of the maximum value in the$F$
array let$now=6$ ( $5$ also works)
$ans$ += F[6]
($ans$ == $18$ )
At this time
$F$ = { $4,2,3,/,3,/$}
$F\left[5\right]$ becomes$3$ , because it is in$6$
There are no pointson the left$F$ value

$X=3$ o'clock
$ans$ += F[1]
($ans$ == $22$)
$F$ = { $/,2,3,/,3,/$}

$X=4$ o'clock
$ans$ += F[3]
($ans$ == $25$)
$F$ = { $/,2,/,/,3,/$}

$X=5$ o'clock
$ans$ += F[5]
($ans$ == $28$)
$F$ = { $/,2,/,/,/,/$}

$X=6$ o'clock
$ans$ += F[2]
($ans$ == $30$)
$F$ = { $/,/,/,/,/,/$ }
The final answer is:

13
18
22
25
28
30

So here comes the question, how to quickly find the maximum value?
We use $maxn[now]$ 表示 $now$ ~ $n$ 中 $F[i]$ 的最大值 ($now<i<=n$)
$The\; role\; of\; maxxn\left[i\right]$ is to maintainthe now nowF [ i ] F[i]on the right side of$n$$o$$w$The maximum value of$F$$[$$i$$]$
thennow nowThe largest a [ i ] a[i]on the left side of$n$$o$$w$How to maintain$a$$[$$i$$]$
? Because$The\; value\; on\; the\; left\; side\; of\; now$ needs to have the operation of deleting the maximum value
, somaxn maxnThe preprocessing operation of$max$$x$$n$
, but priority_queue is a good data structure,
you can use the priority queue$q$ maintains the maximuma [ i ] a[i]the value of$a$$[$$i$$]$

You will find that these operations will not use $F$ array, so there is no need to create this array, let alone change it,
now nowa [ i ] a[i]on the left side of$n$$o$$w$$a\left[i\right]$ will not change, and$maxxn\left[now\right]$ only needs to subtract$s[now]\ast 2$ 即可
$maxn[now]$ = $s[i]\ast 2+a\left[i\right]$ , ( i$ is the maximum value at this time)
so$maxn[now]-s[now]\ast 2$ = $a[i]+s\left[i\right]\ast 2-s\left[now\right]\ast 2$ = $a\left[i\right]+(s[i]-s[now])\ast 2$ is the aboveF [ i ] F[i]value of$F$$[$$i$$]$

Ac Code:

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+10,INF=1e9,mod=INF+7;
int n,now,cnt;
long long ans;
int a[N],s[N];
pair<int,int> maxn[N];
priority_queue<int> q;

int main()
{
    
    
	scanf("%d",&n);
	for(int i = 1;i <= n;i++)
		scanf("%d",s+i);
	for(int i = 1;i <= n;i++)
		scanf("%d",a+i);
	maxn[n+1].second = -INF;
	for(int i = n;i >= 1;i--){
    
    
		maxn[i] = maxn[i+1];
		if(s[i]*2+a[i] >= maxn[i].second)
			maxn[i] = make_pair(i,s[i]*2+a[i]);
	}
	for(int i = 1;i <= n;i++){
    
    
		int l = (q.empty() ? -INF : q.top());
		int r = (now == n ? -INF : maxn[now+1].second);
		if(l >= r-s[now]*2){
    
    
			ans += l;
			q.pop();
			printf("%lld\n",ans);
		}else {
    
    
			ans += r-s[now]*2;
			for(int i = now+1;i <= maxn[now+1].first-1;i++)
				q.push(a[i]);
			now = maxn[now+1].first;
			printf("%lld\n",ans);
		}
	}
	return 0;
}