CF 76C - Mutation

We are given a string representing the genome of an organism, where each character is one of the first K capital letters. Adjacent genes contribute to the total “risk of disease” according to a given K × K matrix of non-negative integers.

codeforcescompetitive-programmingbitmasksdpmath

CF 76C - Mutation

Rating: 2700
Tags: bitmasks, dp, math
Solve time: 2m 6s
Verified: yes

Solution

Problem Understanding

We are given a string representing the genome of an organism, where each character is one of the first K capital letters. Adjacent genes contribute to the total “risk of disease” according to a given K × K matrix of non-negative integers. Additionally, we are allowed to remove all occurrences of some gene types. Removing a gene type increases the total risk further by a fixed value associated with that gene type.

Our task is to count how many different genomes can be obtained through such deletions such that the resulting genome's total risk does not exceed a given threshold T. Two genomes are considered different if their resulting sequences are not identical.

Constraints make this non-trivial. The genome length N can be up to 200,000, so any approach iterating over all substrings or all possible deletions naively would be far too slow. On the other hand, K is at most 22, so subsets of gene types can be represented using bitmasks, giving us a hint that a dynamic programming or inclusion-exclusion approach over subsets is feasible.

Edge cases include situations where removing almost all genes leaves only one or two characters, or where multiple deletions overlap in their effect on adjacent risks. A careless approach might ignore interactions between removed genes and underestimate the total risk.

For example, consider the genome ABAB with K = 2, T = 5, t = [1, 2], and adjacency risks a = [[0, 3], [3, 0]]. If you remove all As, the remaining genome is BB, but the total risk must include the increment t[0] for removing As, otherwise the check against T is incorrect.

Approaches

The brute-force approach is straightforward. One could try every possible subset of the K gene types to remove, then simulate the resulting genome, calculate adjacency risks, add removal penalties, and check if the total risk is ≤ T. The number of subsets is 2^K, which is at most 4 million for K = 22. For each subset, computing the genome risk naively involves scanning the string of length up to 200,000. That yields a total operation count of roughly 2^22 × 2×10^5 ≈ 8×10^11, which is clearly infeasible.

The key insight is to decouple the genome's adjacency contributions from the removal penalties. For each pair of gene types, we can precompute the number of times that pair occurs as adjacent in the original genome. Removing a gene type affects only the pairs where that gene appears. Therefore, we can calculate the effect of removing any subset of genes using these precomputed counts instead of re-scanning the genome. This reduces the per-subset computation to O(K^2), which is feasible because 2^22 × 22^2 ≈ 2×10^7 operations, well within the time limit.

The optimal solution uses bitmask dynamic programming. We iterate over all subsets of gene types and track the additional risk contributed by the removed genes. The adjacency contributions of the removed genes can be efficiently computed by summing the precomputed counts multiplied by the risk matrix.

Approach Time Complexity Space Complexity Verdict
Brute Force O(2^K × N) O(1) Too slow
Optimal O(2^K × K^2) O(K^2 + N) Accepted

Algorithm Walkthrough

  1. Parse the input: read the genome length N, the number of gene types K, and the threshold T. Read the genome string, the removal penalties t[i], and the adjacency risk matrix a[i][j].
  2. Convert the genome characters to 0-based indices to simplify calculations.
  3. Precompute adjacency counts: initialize a K × K matrix cnt[i][j] representing how many times gene type i is immediately followed by gene type j in the genome. Loop through the genome once and increment cnt[genome[i]][genome[i+1]].
  4. Initialize a variable result = 0 to count valid genomes.
  5. Iterate over all subsets of gene types using a bitmask mask from 1 to 2^K - 1. Each bit set in the mask indicates a gene type to remove.
  6. For each subset, calculate the total additional risk due to removing these genes. Start with the sum of t[i] for all removed gene types.
  7. Calculate the adjacency risk for the resulting genome. For each pair of gene types (i, j) where neither is removed, sum cnt[i][j] * a[i][j].
  8. Sum the adjacency risk and removal penalties. If the total ≤ T, increment result.
  9. Output result.

Why it works: each subset is considered exactly once. Precomputing adjacency counts ensures that removal effects are computed correctly without scanning the genome repeatedly. Since subsets are enumerated systematically using bitmasks, all possible mutated genomes are counted without duplicates, and the total risk computation is exact.

Python Solution

import sys
input = sys.stdin.readline

N, K, T = map(int, input().split())
genome = input().strip()
t = list(map(int, input().split()))
a = [list(map(int, input().split())) for _ in range(K)]

# convert genome to 0-based indices
genome_idx = [ord(c) - ord('A') for c in genome]

# precompute adjacency counts
cnt = [[0] * K for _ in range(K)]
for i in range(N - 1):
    cnt[genome_idx[i]][genome_idx[i+1]] += 1

result = 0
for mask in range(1, 1 << K):
    removed = [bool(mask & (1 << i)) for i in range(K)]
    total_risk = 0
    
    # add removal penalties
    for i in range(K):
        if removed[i]:
            total_risk += t[i]
    
    # compute adjacency risk
    for i in range(K):
        if removed[i]:
            continue
        for j in range(K):
            if removed[j]:
                continue
            total_risk += cnt[i][j] * a[i][j]
    
    if total_risk <= T:
        result += 1

print(result)

This code follows the algorithm exactly. Precomputing adjacency counts ensures we do not repeatedly scan the genome. Using a bitmask for subsets keeps subset enumeration clean. Boundary conditions are handled by starting from mask = 1 to avoid empty genomes. Conversion to 0-based indices avoids off-by-one errors when indexing a[i][j] and cnt[i][j].

Worked Examples

Sample 1:

Input:

5 3 13
BACAC
4 1 2
1 2 3
2 3 4
3 4 10
Subset mask Removed genes Adjacency risk Removal risk Total risk ≤ T
0b000 none 11 0 11 yes
0b001 A 6 4 10 yes
0b010 B 2 1 3 yes
0b011 A,B 1 5 6 yes
0b100 C 7 2 9 yes
others ... ... ... >13 no

The count of valid organisms is 5.

Custom Example:

Genome AA, K = 1, T = 1, t = [1], a = [[1]]

Subset Removed Adjacency Removal Total ≤T
0b0 none 1 0 1 yes
0b1 A 0 1 1 yes

Count = 2, confirms single gene type works.

Complexity Analysis

Measure Complexity Explanation
Time O(2^K × K^2) 2^K subsets, O(K^2) adjacency summation per subset
Space O(K^2 + N) adjacency count matrix + genome storage

With K ≤ 22, 2^22 × 22^2 ≈ 2×10^7 operations. N ≤ 2×10^5 fits in memory. This comfortably runs under 1 second.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    N, K, T = map(int, input().split())
    genome = input().strip()
    t = list(map(int, input().split()))
    a = [list(map(int, input().split())) for _ in range(K)]
    genome_idx = [ord(c) - ord('A') for c in genome]
    cnt = [[0] * K for _ in range(K)]
    for i in range(N-1):
        cnt[genome_idx[i