Good fix for this algorithmic puzzle code (USACO)?

Question

I'm a first-year computer science student and I am currently dabbling in some algorithmic competitions. The code below that I made has a flaw that I'm not sure how to fix

Here is the problem statement: http://www.usaco.org/index.php?page=viewproblem2&cpid=811

In the statement, I missed where it said that Farmer John could only switch boots on tiles that both boots can stand on. I tried adding constraints in different places but none seemed to address the problem fully. I don't really see a way to do it without butchering the code

Basically, the problem is that John keeps switching boots on tiles where the new boots can't stand on, and I can't seem to fix it

Here is my code (sorry for the one letter variables):

import java.io.*;
import java.util.*;
public class snowboots {
    static int n,k;
    static int[] field,a,b; //a,b --> strength, distance
    static int pos;
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new FileReader("snowboots.in"));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("snowboots.out")));
        StringTokenizer st = new StringTokenizer(br.readLine());
        n = Integer.parseInt(st.nextToken());
        k = Integer.parseInt(st.nextToken());
        st = new StringTokenizer(br.readLine());
        field = new int[n];
        a = new int[k];
        b = new int[k];
        for (int i = 0; i < n; i++)
            field[i] = Integer.parseInt(st.nextToken());
        for (int i = 0; i < k; i++) {
            st = new StringTokenizer(br.readLine());
            a[i] = Integer.parseInt(st.nextToken());
            b[i] = Integer.parseInt(st.nextToken());
        }

        pw.println(solve());
        pw.close();

    }
    static int solve() {
        pos = 0;
        int i = 0; //which boot are we on?
        while(pos < n-1) {

            while(move(i)); //move with boot i as far as possible

            i++; //use the next boot

        }
        i--;
        return i;
    }
    static boolean move(int c) {
        for (int i = pos+b[c]; i > pos; i--) {
            if (i < n && field[i] <= a[c]) { //snow has to be less than boot strength
                pos = i;
                return true;
            }
        }
        return false;
    }
}

I tried adding a constraint in the "move" method, and one when updating I, but they both are too strict and activate at unwanted times

Is it salvageable?

Answer 1

Yes, it's possible to salvage your solution, by adding an extra for-loop.

What you need to do is, if you find that your previous pair of boots can get you all the way to a tile that's too deep in snow for your next pair, then you need to try "backtracking" to the latest tile that's not too deep. This ends up giving a solution in worst-case O(N·B) time and O(1) extra space.

It may not be obvious why it's OK to backtrack to that tile — after all, just because you can reach a given tile, that doesn't necessarily mean that you were able to reach all the tiles before it — so let me explain a bit why it is OK.

Let maxReachableTileNum be the number (between 1 and N) of the last tile that you were able to reach with your previous boots, and let lastTileNumThatsNotTooDeep be the number (between 1 and N) of the last tile on or before maxReachableTileNum that's not too deeply snow-covered for your next pair. (We know that there is such a tile, because tile #1 has no snow at all, so if nothing else we know that we can backtrack to the very beginning.) Now, since we were able to get to maxReachableTileNum, then some previous boot must have either stepped on lastTileNumThatsNotTooDeep (in which case, no problem, it's reachable) or skipped over it to some later tile (on or before maxReachableTileNum). But that later tile must be deeper than lastTileNumThatsNotTooDeep (because that later tile's depth is greater than s_{currentBootNum}, which is at least at great as the depth of lastTileNumThatsNotTooDeep), which means that the boot that skipped over lastTileNumThatsNotTooDeep certainly could have stepped on lastTileNumThatsNotTooDeep instead: it would have meant taking a shorter step (OK) onto a less-deeply-covered tile (OK) than what it actually did. So, either way, we know that lastTileNumThatsNotTooDeep was reachable. So it's safe for us to try backtracking to lastTileNumThatsNotTooDeep. (Note: the below code uses the name reachableTileNum instead of lastTileNumThatsNotTooDeep, because it continues to use the reachableTileNum variable for searching forward to find reachable tiles.)

However, we still have to hold onto the previous maxReachableTileNum: backtracking might turn out not to be helpful (because it may not let us make any further forward progress than we already have), in which case we'll just discard these boots, and move on to the next pair, with maxReachableTileNum at its previous value.

So, overall, we have this:

public static int solve(
    final int[] tileSnowDepths,           // tileSnowDepths[0] is f_1
    final int[] bootAllowedDepths,        // bootAllowedDepths[0] is s_1
    final int[] bootAllowedTilesPerStep   // bootAllowedTilesPerStep[0] is d_1
) {
    final int numTiles = tileSnowDepths.length;
    final int numBoots = bootAllowedDepths.length;
    assert numBoots == bootAllowedTilesPerStep.length;

    int maxReachableTileNum = 1; // can reach tile #1 even without boots

    for (int bootNum = 1; bootNum <= numBoots; ++bootNum) {
        final int allowedDepth = bootAllowedDepths[bootNum-1];
        final int allowedTilesPerStep = bootAllowedTilesPerStep[bootNum-1];

        // Find the starting-point for this boot -- ideally the last tile
        // reachable so far, but may need to "backtrack" if that tile is too
        // deep; see explanation above of why it's safe to assume that we
        // can backtrack to the latest not-too-deep tile:

        int reachableTileNum = maxReachableTileNum;
        while (tileSnowDepths[reachableTileNum-1] > allowedDepth) {
            --reachableTileNum;
        }

        // Now see how far we can go, updating both maxReachableTileNum and
        // reachableTileNum when we successfully reach new tiles:

        for (int tileNumToTry = maxReachableTileNum + 1;
             tileNumToTry <= numTiles
                 && tileNumToTry <= reachableTileNum + allowedTilesPerStep;
             ++tileNumToTry
        ) {
            if (tileSnowDepths[tileNumToTry-1] <= allowedDepth) {
                maxReachableTileNum = reachableTileNum = tileNumToTry;
            }
        }

        // If we've made it to the last tile, then yay, we're done:

        if (maxReachableTileNum == numTiles) {
            return bootNum - 1; // had to discard this many boots to get here
        }
    }

    throw new IllegalArgumentException("Couldn't reach last tile with any boot");
}

(I tested this on USACO's example data, and it returned 2, as expected.)

This can potentially be optimized further, e.g. with logic to skip pairs of boots that clearly aren't helpful (because they're neither stronger nor more agile than the previous successful pair), or with an extra data structure to keep track of the positions of latest minima (to optimize the backtracking process), or with logic to avoid backtracking further than is conceivably useful; but given that N·B ≤ 250² = 62,500, I don't think any such optimizations are warranted.

---

Edited to add (2019-02-23): I've thought about this further, and it occurs to me that it's actually possible to write a solution in worst-case O(N + B log N) time (which is asymptotically better than O(N·B)) and O(N) extra space. But it's much more complicated; it involves three extra data-structures (one to keep track of the positions of latest minima, to allow backtracking in O(log N) time; one to keep track of the positions of future minima, to allow checking in O(log N) time if the backtracking is actually helpful (and if so to move forward to the relevant minimum); and one to maintain the necessary forward-looking information in order to let the second one be maintained in amortized O(1) time). It's also complicated to explain why that solution is guaranteed to be within O(N + B log N) time (because it involves a lot of amortized analysis, and making a minor change that might seem like an optimization — e.g., replacing a linear search with a binary search — can break the analysis and actually increase the worst-case time complexity. Since N and B are both known to be at most 250, I don't think all the complication is worth it.

Answer 2

One way which to solve it using BFS. You may refer below code for details. Hope this helps.

import java.util.*;
import java.io.*;

public class SnowBoots {

    public static int n;
    public static int[] deep;
    public static int nBoots;
    public static Boot[] boots;

    public static void main(String[] args) throws Exception {

        // Read the grid.
        Scanner stdin = new Scanner(new File("snowboots.in"));

        // Read in all of the input.
        n = stdin.nextInt();
        nBoots = stdin.nextInt();
        deep = new int[n];
        for (int i = 0; i < n; ++i) {
            deep[i] = stdin.nextInt();
        }
        boots = new Boot[nBoots];
        for (int i = 0; i < nBoots; ++i) {
            int d = stdin.nextInt();
            int s = stdin.nextInt();
            boots[i] = new boot(d, s);
        }

        PrintWriter out = new PrintWriter(new FileWriter("snowboots.out"));
        out.println(bfs());
        out.close();
        stdin.close();
    }

    // Breadth First Search Algorithm [https://en.wikipedia.org/wiki/Breadth-first_search]
    public static int bfs() {

        // These are all valid states.
        boolean[][] used = new boolean[n][nBoots];
        Arrays.fill(used[0], true);

        // Put each of these states into the queue.
        LinkedList<Integer> q = new LinkedList<Integer>();
        for (int i = 0; i < nBoots; ++i) {
            q.offer(i);
        }

        // Usual bfs.
        while (q.size() > 0) {

            int cur = q.poll();
            int step = cur / nBoots;
            int bNum = cur % nBoots;

            // Try stepping with this boot...
            for (int i = 1; ((step + i) < n) && (i <= boots[bNum].maxStep); ++i) {
                if ((deep[step+i] <= boots[bNum].depth) && !used[step+i][bNum]) {
                    q.offer(nBoots * (step + i) + bNum);
                    used[step + i][bNum] = true;
                }
            }

            // Try switching to another boot.
            for (int i = bNum + 1; i < nBoots; ++i) {
                if ((boots[i].depth >= deep[step]) && !used[step][i]) {
                    q.offer(nBoots * step + i);
                    used[step][i] = true;
                }
            }
        }

        // Find the earliest boot that got us here.
        for (int i = 0; i < nBoots; ++i) {
            if (used[n - 1][i]) {
                return i;
            }
        }

        // Should never get here.
        return -1;
    }
}

class Boot {

    public int depth;
    public int maxStep;

    public Boot(int depth, int maxStep) {
        this.depth = depth;
        this.maxStep = maxStep;
    }
}

Answer 3

You can solve this problem by Dynamic Programming. You can see the concept in this link (Just read the Computer programming part). It has following two steps.

• First solve the problem recursively.
• Memoize the states.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mx 100005
#define mod 1000000007
int n, b;
int f[333], s[333], d[333];
int dp[251][251];
int rec(int snowPos, int bootPos)
{
    if(snowPos == n-1){
        return 0;

    int &ret = dp[snowPos][bootPos];
    if(ret != -1) return ret;
    ret = 1000000007;
    for(int i = bootPos+1; i<b; i++)
    {
        if(s[i] >= f[snowPos]){
            ret = min(ret, i - bootPos + rec(snowPos, i));
        }
    }
    for(int i = 1; i<=d[bootPos] && snowPos+i < n; i++){
        if(f[snowPos + i] <= s[bootPos]){
            ret = min(ret, rec(snowPos+i, bootPos));
        }
    }
    return ret;
}
int main()
{
    freopen("snowboots.in", "r", stdin);
    freopen("snowboots.out", "w", stdout);
    scanf("%d %d", &n, &b);
    for(int i = 0; i<n; i++)
        scanf("%d", &f[i]);
    for(int i = 0; i<b; i++){
        scanf("%d %d", &s[i], &d[i]);
    }
    memset(dp, -1, sizeof dp);
    printf("%d\n", rec(0, 0));

    return 0;
}

This is my solution to this problem (in C++). This is just a recursion. As problem says,

• you can change boot, Or
• you can do a jump by current boot.

Memoization part is done by the 2-Dimensional array dp[][].