# My Editorial to TopCoder SRM 638 Div 2

SRM 638 took place in the middle of the night in my time zone, so I didn't participate.

While I haven't read problems from my division (Div 1) yet, I decided to write an editorial for a whole Div 2 problem set.

One interesting thing is that first two problems can be solve very very fast in python, so having an ability to code in python can be a great advantage during live contest.

1000 problem was solved only by 3 users during the contest so I will give you a really simple method to implement it.

## 250 Problem // NamingConvention

#### Problem statement

This is really simple. You are given a string consisting of lowercase letters a-z and underscores '_', where underscores divide consecutive words and there is at least one letter between two underscores. The task is to convert the given string into camel case notation, i.e. first word is written in lower case and every other word starts with a capital letter while other letters are lower case.

The full problem statement is available here: http://community.topcoder.com/stat?c=problem_statement&pm=13521&rd=16081

#### Solution

You can split the string by underscores and capitalize the first letter of each word starting from the second one. Print resulting words one by one. In python this is so simple:

class NamingConvention: def toCamelCase(self, variableName): s = variableName.split('_') return "".join([s[0]] + map(lambda x: x.title(), s[1:]))

The title() method of the class string class is very useful here. The solution has of course linear time complexity.

## 500 Problem // NarrowPassage2Easy

#### Problem statement

You are given an array $A$ of size at most $6$, where every element is a positive integer not greater than $1000$. You are also given a positive integer $K$ not greater than $1000$.

The task is to count the number of different orders of $A$ (like permutations, but two elements with same value are considered different) for which there are no elements $a_i$, $a_j$ such that $a_i + a_j > +K$ and $a_j$ was before $a_i$ in the original array.

The full problem statement is available here: http://community.topcoder.com/stat?c=problem_statement&pm=13295&rd=16081

#### Solution

Since the array has at most 6 elements, we can consider every possible ordering and for each one check if there is a pair of elements, for which the condition from the problem statement is invalid. If there is no such pair, then we count this ordering as a valid and at the end we return the number of valid orderings. Again python makes it so easy to implement:

class NarrowPassage2Easy: def count(self, A, K): from itertools import permutations cnt = 0 n = len(A) for p in permutations(enumerate(A)): fail = 0 for i in range(n): for j in range(i + 1, n): if p[i][0] &gt; p[j][0] and p[i][1] + p[j][1] &gt; K: fail = 1 if not fail: cnt += 1 return cnt

If you haven't heard about enumerate() function this is the time to read about it. Time complexity of this method is $O(n^2 \cdot n!)$ where $n$ is the size of the input array and since $n \leq 6$ it is enough here.

## 1000 Problem // CandleTimerEasy

#### Problem statement

You are given a tree consisting of $n$ nodes. Each edge in the tree has a positive length of at most $1000$ units.

You can ignite any subset of leaves in the tree. When a node $v$ is ignited, all edges adjacent to it starts burning, and if the fire meets another node $u$, it becomes ignited and the process continues. The whole process lasts until all edges are burned.

Edges burn at a uniform rate. This means that if you ignite an edge of length $L$ at one end, it will burn in $L$ units of time, but if you place fire at its both ends, it will burn quicker (for example, if you place it in the same time, it will burn after $L/2$ units of time).

The tree is burned if all its edges are burned.

The task is count the number of different times in which the tree can be burned completely. These times may differ if you ignite different subsets of leaves at the beginning.

The full problem statement is available here: http://community.topcoder.com/stat?c=problem_statement&pm=13519&rd=16081

#### Solution

Since $n$ is at most $20$ here, we can try to ignite all subsets of leaves and for each one count the time in which the tree will burn completely.

At first I thought about simulating the process of burning, but it can be quite complex to implement, so I developed a solution very easy to code.

Let $d[v][u]$ be the time in which $v$ is ignited when you ignite only $u$at the beginning. In other words, $d[v][u]$ is a distance between $v$ and $u$ where $u$ is a leaf in the tree. You can compute $d$ table using dfs for example.

Next we iterate over all subset of ignited leaves. I will show how to count the burning time for a given subset $S$ of size $k$.

Let $f[v]$ be the time when a node $v$ is ignited. We ignite at the beginning all leaves from $S$. You can thing of this process as of $k$ different fire flames spreading on the tree. Node $v$ will be ignited by one of these flames. Which one? The one which comes to $v$ first and since we have our $d$ table computed, we can compute $f[v] = \min\limits_{u \in S} d[v][u]$.

Now we know $f[v]$ for any node in the tree. Since the tree is burned when all its edges are burned, we iterate over all edges and for each edge $(v, u)$ we find the time after it becomes completely burned based on values $f[v]$ and $f[u]$.

Time complexity of this method is $O(2^L \cdot L \cdot n)$ where $L$ is the number of leaves in the tree and $n$ is the number of all nodes in it.

I attached my C++ code below:

const int MAX_N = 20; vi g[MAX_N]; int w[MAX_N][MAX_N]; vi leaves; int fire1[MAX_N]; int fire2[MAX_N][MAX_N]; void dfs(int v, int p, int len, int root) { fire2[v][root] = len; FOR(i, SZ(g[v])) { int u = g[v][i]; if(u == p) continue; dfs(u, v, len + w[v][u], root); } } class CandleTimerEasy { public: int differentTime(vector A, vector B, vector len) { int n = SZ(A) + 1; FOR(i, n) g[i].clear(); leaves.clear(); FOR(i, n - 1) { g[A[i]].pb(B[i]); g[B[i]].pb(A[i]); w[B[i]][A[i]] = len[i]; w[A[i]][B[i]] = len[i]; } FOR(i, n) if(SZ(g[i]) == 1) leaves.pb(i); set ans; FOR(i, n) dfs(i, i, 0, i); for(int b = 1; b &lt; (1 &lt;&lt; SZ(leaves)); ++b) { int k = b; int c = 0; vector vs; while(k &gt; 0) { int v = leaves[c]; if(k &amp; 1) vs.pb(v); k /= 2; ++c; } FOR(i, n) fire1[i] = INF; FOR(v, n) { FOR(i, SZ(vs)) { int r = vs[i]; REMIN(fire1[v], fire2[v][r]); } } double res = 0; FOR(v, n) { FOR(i, SZ(g[v])) { int u = g[v][i]; double burned = min(fire1[v], fire1[u]) + w[v][u]; int dif = abs(fire1[v] - fire1[u]); REMIN(burned, min(fire1[v], fire1[u]) + dif + (w[v][u] - dif) / 2.0); REMAX(res, burned); } } ans.insert(res); } return SZ(ans); } };

# TopCoder / SRM 622 div1 250

SRM 622 took place 10 hours ago, but I didn't participate because it was in the middle of the night in Poland. Here's a quick post on the 250 problem.

## 250 Problem

You're given an integer $T$ and a full weighted directed graph of at most $N \leq 50$ vertices, where $w_{i,j}$ is the weight of the edge connecting vertex $i$ with vertex $j$. The task is to count a sum of weights of overloaded edges. An edge $e$ is overloaded if and only if there are at least $T$ pairs of vertices $(v, u)$ such that $e$ lies on any shortest path between $v$ and $u$. The full problem statement is here SRM 622 div1 250.

The simplest method that came to my mind is to first compute the lengths of shortest paths between all pairs of vertices. Let $d_{v, u}$ denote the distance between $v$ and $u$. During the contest, the crucial thing here is to use the Floyd-Warshall algorithm rather than for example Dijkstra, not because it's faster, but because it's significantly easier to code.

It remains to decide if an edge $e = (v, u)$ lies on any shortest path between given vertices $s$ and $t$. Of course we cannot generate all shortest paths from $s$ to $t$, because the number of such paths can be $\Omega(2^n)$. The most clever method to check it is to observe that:

$e = (v, u)$ lies on any shortest path from $s$ to $t$ if and only if

$d_{s, t} = d_{s, v} + w_{v, u} + d_{u, t}$

This can be checked in $O(1)$ time because we've just computed distances for all pairs of vertices.

The time complexity of my algorithm is dominated by the cost of doing the above check for the whole graph. For each pair of verices, we iterate over all edges and do a constant time work which results in $O(n^4)$.

Here's my code:

const int MAX_N = 55;

int d[MAX_N][MAX_N];
int cnt[MAX_N][MAX_N];
class BuildingRoutes {
public:
int build(vector w, int T) {
int n = w.size();
FOR(i, n) FOR(j, n) d[i][j] = w[i][j] - '0';
FOR(k, n)
{
FOR(i, n)
{
FOR(j, n)
{
REMIN(d[i][j], d[i][k] + d[k][j]);
}
}
}
FOR(s, n)
{
FOR(t, n)
{
FOR(i, n)
{
FOR(j, n)
{
if(d[s][t] == d[s][i] + (w[i][j] - '0') + d[j][t])
cnt[i][j]++;
}
}
}
}
int res = 0;
FOR(i, n)
{
FOR(j, n)
{
if(cnt[i][j] >= T) res += w[i][j] - '0';
}
}
return res;
}
};