Challenging problems require innovative approaches to solve efficiently.

Short Colorful Strip

Given that n equals m, the final configuration must be a permutation of n.

Key observations:

1. When coloring an interval, the smallest color within that interval is always colored first.
2. The coloring operation requires all points in the covered interval to have the same color.

For the initial step, we locate the position p of the color 1.

The coloring interval [l, r] must satisfy l ≤ p ≤ r.

This allows us to partition the solution into four segments: [1, l], [l+1, p-1], [p+1, r], and [r+1, n]

The solution can be expressed as: ans = Σ(l=1 to p-1) f[1][l] × f[l+1][p-1] × Σ(r=p to n) f[p+1][r] × f[r+1][n]

This approach can be applied to any queried interrval, leading to a clear solution strategy.

const int N = 505;
const int mod = 1000000007;
int size, operations;
int sequence[N];
int dp[N][N];

int solve(int left, int right) {
    if (dp[left][right] != -1) return dp[left][right];
    if (left >= right) return dp[left][right] = 1;
    
    int pos = left;
    for (int i = left; i <= right; i++) 
        if (sequence[i] < sequence[pos]) pos = i;
    
    long long left_sum = 0, right_sum = 0;
    
    for (int i = left; i <= pos; i++) 
        left_sum = (left_sum + solve(left, i-1) * solve(i, pos-1)) % mod;
    
    for (int i = pos; i <= right; i++) 
        right_sum = (right_sum + solve(pos+1, i) * solve(i+1, right)) % mod;
    
    return dp[left][right] = (left_sum * right_sum) % mod;
}

int main() {
    memset(dp, -1, sizeof dp);
    size = read(); operations = read();
    for (int i = 1; i <= operations; i++) 
        sequence[i] = read();
    
    cout << solve(1, operations);
    return 0;
}

Long Colorful Strip

A similar approach can be applied, but for each color, we need to find its leftmost and rightmost positions. By enumerating l and r, we can divide the current interval into five segments and perform interval DP.

However, with m ≤ 10^6, this approach is computationally infeasible.

To optimize, we consider compressing consecutive identical colors into a single point. If the compressed length exceeds 2n, we can immediately determine the solution is invalid.

The complexity is O(m^3), but it's acceptable for the problem constraints.

const int N = 1505;
const int mod = 1000000007;
int size, operations;
int sequence[N], compressed_length;
int dp[N][N];
int left_bound[N], right_bound[N];

int solve(int left, int right) {
    if (dp[left][right] != -1) return dp[left][right];
    if (left > right) return 1;
    
    int &result = dp[left][right];
    result = 0;
    
    int min_val = sequence[left];
    for (int j = left; j <= right; j++) 
        min_val = min(min_val, sequence[j]);
    
    int last_pos = -1;
    int middle_product = 1;
    
    for (int j = left; j <= right; j++) {
        if (sequence[j] == min_val) {
            if (last_pos != -1) 
                middle_product = middle_product * solve(last_pos+1, j-1) % mod;
            last_pos = j;
        }
    }
    
    long long left_sum = 0;
    for (int j = left; j <= left_bound[min_val]; j++) 
        left_sum = (left_sum + solve(left, j-1) * solve(j, left_bound[min_val]-1)) % mod;
    
    long long right_sum = 0;
    for (int j = right_bound[min_val]; j <= right; j++) 
        right_sum = (right_sum + solve(right_bound[min_val]+1, j) * solve(j+1, right)) % mod;
    
    return result = middle_product * left_sum % mod * right_sum % mod;
}

int main() {
    memset(dp, -1, sizeof dp);
    size = read(); operations = read();
    
    int current, prev = 0;
    for (int i = 1; i <= operations; i++) {
        current = read();
        if (current != prev) {
            sequence[++compressed_length] = current;
            prev = current;
        }
    }
    
    if (compressed_length > 2 * size) {
        puts("0");
        return 0;
    }
    
    for (int i = 1; i <= size; i++) {
        int l = compressed_length + 1, r = 0;
        for (int j = 1; j <= compressed_length; j++) {
            if (sequence[j] == i) {
                l = min(l, j);
                r = max(r, j);
            }
        }
        
        for (int j = l; j <= r; j++) {
            if (sequence[j] < i) {
                puts("0");
                return 0;
            }
        }
        
        left_bound[i] = l;
        right_bound[i] = r;
    }
    
    cout << solve(1, compressed_length);
    return 0;
}

Palindrome

This is a classic problem with an important property:

The longest palindromic subsequence equals the longest common subsequence (LCS) between the original string and its reverse.

However, computing LCS has O(n^2) complexity.

Since all characters are lowercase letters, if the string length exceeds 2600, there must exist a valid subsequence where all characters are the same.

Otherwise, we directly compute the LCS.

const int MAX_LEN = 2600;
const int MIN_OUTPUT = 100;
const int MAX_OUTPUT = 5050;
int size;
string input, reversed;
int dp[MAX_LEN+1][MAX_LEN+1];
char result[MAX_OUTPUT], doubled_result[MAX_OUTPUT];

int main() {
    cin >> input;
    size = input.size();
    reversed = input;
    reverse(reversed.begin(), reversed.end());
    
    input = " " + input;
    reversed = " " + reversed;
    
    if (size <= MAX_LEN) {
        for (int i = 1; i <= size; i++) {
            for (int j = 1; j <= size; j++) {
                if (input[i] == reversed[j]) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                } else {
                    dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        
        int i = size, j = size;
        int pos = dp[size][size];
        while (pos > 0) {
            if (input[i] == reversed[j]) {
                result[pos] = input[i];
                i--; j--; pos--;
            } else {
                if (dp[i][j] == dp[i-1][j]) i--;
                else j--;
            }
        }
        
        if (dp[size][size] < MIN_OUTPUT) {
            printf("%s", result + 1);
            return 0;
        } else {
            int len = 0;
            for (int i = 1; i <= MIN_OUTPUT/2; i++) 
                doubled_result[++len] = result[i];
            
            for (int i = dp[size][size] - MIN_OUTPUT/2; i >= 1; i--) 
                doubled_result[++len] = result[i];
            
            printf("%s", doubled_result + 1);
        }
    } else {
        for (char c = 'a'; c <= 'z'; c++) {
            int count = 0;
            for (int i = 1; i <= size; i++) {
                if (input[i] == c) count++;
            }
            
            if (count >= MIN_OUTPUT) {
                for (int i = 1; i <= MIN_OUTPUT; i++) 
                    putchar(c);
                return 0;
            }
        }
    }
    return 0;
}

Constrained Tree

We already know the preorder traversal sequence of this binary tree.

To determine if a node belongs to the left or right subtree, we can compute the minimum and maximum values for the left and right subtrees of each node.

Then we can recursively construct the tree.

const int N = 1005;
int size, constraints;
int left_max[N], right_max[N], left_min[N], right_min[N];
int node_count = 0;
bool valid = true;
vector<int> traversal;

void build_tree(int left, int right) {
    if (!valid) return;
    
    node_count++;
    if (left_max[left]) {
        if (left_min[left] <= node_count) valid = false;
        build_tree(node_count+1, left_max[left]);
    }
    
    traversal.push_back(left);
    
    if (right_max[left]) {
        if (right_min[left] <= node_count) valid = false;
        build_tree(node_count+1, max(right_max[left], right));
    } 
    else if (node_count < right) {
        build_tree(node_count+1, right);
    }
}

int main() {
    size = read(); constraints = read();
    
    for (int i = 1; i <= size; i++) {
        left_min[i] = size + 1;
        right_min[i] = size + 1;
    }
    
    int x, y;
    char direction[10];
    
    for (int i = 1; i <= constraints; i++) {
        x = read(); y = read();
        if (x >= y) {
            puts("IMPOSSIBLE");
            return 0;
        }
        
        scanf("%s", direction+1);
        
        if (direction[1] == 'L') {
            left_max[x] = max(left_max[x], y);
            left_min[x] = min(left_min[x], y);
        } else {
            right_max[x] = max(right_max[x], y);
            right_min[x] = min(right_min[x], y);
        }
    }
    
    build_tree(1, size);
    
    if (!valid) {
        puts("IMPOSSIBLE");
    } else {
        for (auto node : traversal) {
            cout << node << " ";
        }
    }
    
    return 0;
}