Problem: Link

Submission: Link $$$O(n)$$$, 92ms!

Part1: Hint

By looking at the $$$4$$$ test cases, we find that answers for $$$b[i] = 1$$$ are the same.

Part2: The greedy strategy (for each index $$$i$$$)

First, we define some macros for convenience:

The previous index of $$$x$$$: #define pr(x) ((x) == 1 ? n : (x) - 1)

The next index of $$$x$$$: #define ne(x) ((x) == n ? 1 : (x) + 1)

If $$$b[x]==1$$$, rt(x) denotes the place of the next $$$1$$$ in the loop: rt(x). For example, let $$$b$$$ be an $$$1$$$-indexed array $$$[2, 1, 2, 1]$$$, then $$$rt(2)==4$$$ and $$$rt(4)==2$$$. Note that $$$rt(i)==i$$$ could happen, for example, $$$b==[2, 1, 2]$$$, then $$$rt(2)==2$$$. rt could be pre-calculated using a stack.

get(i, j): The gas need to go from $$$i$$$ to $$$j$$$, both inclusive and $$$1$$$-indexed. For example, if $$$a=[1, 2, 3, 4]$$$, then $$$get(1, 2)==a[1]+a[2]==1+2==3$$$ and $$$get(4, 2)==a[4]+a[1]+a[2]==4+1+2==7$$$. The $$$get$$$ function could be pre-calculated using the prefix sum.

We need to maintain an array $$$c$$$ in the algorithm. $$$c[i]$$$ denotes the minimum burbles we need to pay from the previous place $$$j$$$ where $$$b[j]==1$$$ to $$$i$$$. $$$j$$$ may be equal to $$$1$$$. For example, if $$$b==[1, 2, 1]$$$, then $$$c[3]$$$ denotes the minimum burbles we need to pay from $$$3$$$ to $$$3$$$, which is $$$a[3]$$$. $$$c[2]$$$ denotes the minimum burbles we need to pay from $$$1$$$ to $$$2$$$. Note that this $$$c$$$ is well-defined iff there's at least one $$$i$$$ such that $$$b[i]==1$$$. So we just need to output the case where all $$$b[i]$$$ equal to $$$2$$$.

//The Initialpoint
int i = initialpoint;
while(true){
    (1) If you are at the destination, which is the same as the initialpoint, goto finish.

    (2) When you stay at i where b[i]==2, just pay 2a[i] for a[i] gas that enables you to move from i to ne(i). let i = ne(i);

    (3) When you stay at i where b[i]==1, set initialized2 = true and curi = i, and do the following inner while loop. Also, we need to calculate the array c.
    while(initialized2 || curi != rt(i)){ 
        //rt(i): The next i in this trip with b[i]==1
        //Why do we need this initialized2? Because curi before the loop is i, which may be equal to rt(i) at the beginning, in this case we still need to go into this loop!
        if(curi == initialpoint) goto finish;
        initialized2 = false;
        if(get(i, curi) <= k) c[curi] = get(i, curi);
        else c[curi] = k + 2*(get(i, curi) - k); //By k gas at i using k burbles and the left get(i, curi) - k gas using 2*(get(i, curi) - k) burbles.
        curi = ne(curi);
    }
}
finish:

You may worry about the line if(get(i, curi) <= k) c[curi] = get(i, curi);. In this line, if get(i, curi) <= k, we just buy get(i, curi) instead of fill all $$$k$$$ gas into the tank. You may think it would be better if we grasp the chance (that we are at some good place that $$$b[i]==1$$$) to make our tank full. But, our algorithm is correct. Whenever our algorithm is forced to buy one gas using two burbles, every other strategy needs to. If we have not met an $$$i$$$ such that $$$b[i]==1$$$ in our trip, then we can only buy one gas using two burbles, no matter what our strategy is. Otherwise, if we are forced to buy one gas using two burbles at some place $$$i$$$ where $$$b[i]==2$$$, let $$$j$$$ be the last $$$1$$$ we meet ($$$b[j]==1$$$). According to our algorithm, $$$get(j, i) > k$$$, which means our algorithm already makes the tank full at place $$$j$$$. Even if that, it still face the lack of gas issue at $$$i$$$ and has to pay for more expensive gas. For other strategy, the gas amount is at most $$$k$$$ at place $$$j$$$, so it must also face the lack of gas issue at $$$i$$$, and has to pay for more expensive gas also!