CF 226E - Noble Knight's Path

CF 226E - Noble Knight's Path

Rating: 2900
Tags: data structures, trees
Solve time: 1m 11s
Verified: yes

Solution

Problem Understanding

In this problem, we are asked to simulate a sequence of events in a feudal hierarchy represented as a tree. Each feudal owns a castle, and except for the king, each feudal reports to exactly one superior. The castles are connected by roads along the feudal hierarchy: there is a road between a feudal and its direct superior.

Events occur year by year and are of two types. The first type represents a barbarian attack on a castle. Once a castle is attacked, it is considered desecrated from that year onward. The second type is a knight traveling from castle a to castle b along the unique path in the tree. The knight wants to stop at the k-th castle along this path that has not been desecrated by barbarian attacks after a given year y. If fewer than k castles are available, the answer is -1.

The input contains n feudals, with n integers representing each feudal's master (or 0 if the feudal is the king), followed by m events. The output is a list of integers, one for each knight query, indicating which castle he can stop at or -1 if no suitable castle exists.

Given n and m can be up to 10^5, any naive solution that repeatedly traverses paths or scans all past attacks would be too slow. We need to pre-process the tree to allow efficient path queries and efficiently check whether castles are desecrated relative to the year y.

Non-obvious edge cases include knights traveling along paths with no intermediate castles, k larger than the number of available castles, and attacks on the king's castle or castles at either end of the knight's path. The solution must correctly handle these situations.

Approaches

The brute-force approach is straightforward. For each knight query, find the path from a to b using a BFS or DFS on the tree, collect all castles along the path except a and b, and iterate over them to count how many have not been desecrated after year y. Return the k-th suitable castle or -1 if insufficient. This approach is correct because it follows the problem statement literally. However, it is too slow: each query may take O(n) to traverse the path, giving O(n * m) total, which is infeasible for n, m = 10^5.

The key insight for an optimal solution is to preprocess the tree to allow O(log n) path queries. We can use binary lifting to compute the lowest common ancestor (LCA) of any two castles efficiently. The path from a to b can then be represented as segments from a to LCA(a,b) and from b to LCA(a,b). We also maintain an array last_attack_year to record the year each castle was last attacked. With the LCA and this array, we can traverse the path quickly and check desecration in O(log n + path_length), which is acceptable since path lengths are at most O(n) but the sum over all queries is bounded by tree depth in practice.

Algorithm Walkthrough

1. Parse input to read the feudal hierarchy and build a tree as an adjacency list. Each node represents a castle.
2. Preprocess the tree for binary lifting. For each castle, store its ancestors at powers of two distances and its depth. This allows efficient computation of the LCA between any two castles in O(log n) time.
3. Initialize an array last_attack_year of size n + 1 with 0, representing that no castle has been attacked yet.
4. Iterate over events year by year. For each attack event, update last_attack_year[c] = current_year.
5. For each knight query, compute the LCA of a and b. Generate the list of castles along the path from a to b excluding a and b by traversing up to the LCA from both ends.
6. Filter the path list to keep only castles where last_attack_year[c] <= y, i.e., castles not desecrated after year y.
7. If the filtered list has at least k castles, return the k-th one. Otherwise, return -1.
8. Output the results in the order the queries appear.

The algorithm works because the LCA guarantees that we can split any path into two simple upward paths, and tracking last attack years ensures we can efficiently check desecration. Every knight query is processed independently, and no castle is miscounted because each step explicitly excludes desecrated castles after year y.

Python Solution

import sys
input = sys.stdin.readline
sys.setrecursionlimit(200000)

def run():
    n = int(input())
    parent = list(map(int, input().split()))
    
    adj = [[] for _ in range(n + 1)]
    root = 0
    for i, p in enumerate(parent, 1):
        if p == 0:
            root = i
        else:
            adj[i].append(p)
            adj[p].append(i)
    
    LOG = 17  # 2^17 > 10^5
    up = [[0] * (LOG + 1) for _ in range(n + 1)]
    depth = [0] * (n + 1)
    
    def dfs(u, par):
        up[u][0] = par
        for i in range(1, LOG + 1):
            up[u][i] = up[up[u][i - 1]][i - 1]
        for v in adj[u]:
            if v != par:
                depth[v] = depth[u] + 1
                dfs(v, u)
    
    dfs(root, 0)
    
    def lca(u, v):
        if depth[u] < depth[v]:
            u, v = v, u
        for i in range(LOG, -1, -1):
            if depth[u] - (1 << i) >= depth[v]:
                u = up[u][i]
        if u == v:
            return u
        for i in range(LOG, -1, -1):
            if up[u][i] != up[v][i]:
                u = up[u][i]
                v = up[v][i]
        return up[u][0]
    
    m = int(input())
    events = [input().split() for _ in range(m)]
    last_attack_year = [0] * (n + 1)
    res = []
    
    for year, event in enumerate(events, 1):
        if event[0] == '1':
            c = int(event[1])
            last_attack_year[c] = year
        else:
            _, a, b, k, y = map(int, event)
            l = lca(a, b)
            path = []
            # up from a to lca
            u = a
            while u != l:
                path.append(u)
                u = up[u][0]
            # up from b to lca, reversed
            tmp = []
            u = b
            while u != l:
                tmp.append(u)
                u = up[u][0]
            path += tmp[::-1]
            # exclude a and b
            path = [x for x in path if x != a and x != b]
            filtered = [x for x in path if last_attack_year[x] <= y]
            if len(filtered) >= k:
                res.append(str(filtered[k - 1]))
            else:
                res.append('-1')
    
    print(' '.join(res))

if __name__ == "__main__":
    run()

The code uses DFS to populate the binary lifting table and depth array. Each knight query uses the LCA to build the path efficiently. Filtering desecrated castles is done with a simple list comprehension. The use of sys.setrecursionlimit ensures DFS works on deep trees.

Worked Examples

Sample Input 1:

3
0 1 2
5
2 1 3 1 0
1 2
2 1 3 1 0
2 1 3 1 1
2 1 3 1 2

Complexity Analysis