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Adiciona versão simplificada da DFS para achar diâmetro #105

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Adiciona versão simplificada da DFS para achar diâmetro #105

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ac011ea
Adiciona versão simplificada da DFS para achar diâmetro
iagorrr Nov 1, 2023
a9fa111
Corrige erros de digitação e renomeia variáveis
iagorrr Nov 1, 2023
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66 changes: 66 additions & 0 deletions Grafos/slides/arvores_diametro/codes/dp2.cpp
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Original file line number Diff line number Diff line change
@@ -0,0 +1,66 @@
#include <bits/stdc++.h>

using namespace std;
using ii = pair<int, int>;

const int MAX { 2 * 1'00'000 + 1 };
vector<int> adj[MAX];
int to_leaf[MAX], max_length[MAX];

void dfs(int u, int p)
{
int ds1, ds2;
ds1 = ds2 = -1;

for (auto v : adj[u])
{
if (v == p)
continue;

if (ds1 < ds2) swap(ds1, ds2);

dfs(v, u);

ds2 = max(ds2, to_leaf[v]);
}


to_leaf[u] = max(ds1, ds2) + 1;
max_length[u] = 2 + ds1 + ds2;
}

int diameter(int root, int N)
{
dfs(root, 0);

int d = 0;

for (int u = 1; u <= N; ++u)
d = max(d, max_length[u]);

return d;
}

int main()
{
vector<ii> edges { ii(1, 7), ii(3, 7), ii(7, 4), ii(4, 2),
ii(4, 5), ii(5, 6) };

for (const auto& [u, v] : edges) {
adj[u].push_back(v);
adj[v].push_back(u);
}

// 4
cout << diameter(4, 7) << endl;

// 0 0 0 2 1 0 1
for (int u = 1; u <= 7; ++u)
cout << to_leaf[u] << (u == 7 ? '\n' : ' ');

// 0 0 0 4 1 0 2
for (int u = 1; u <= 7; ++u)
cout << max_length[u] << (u == 7 ? '\n' : ' ');

return 0;
}
149 changes: 149 additions & 0 deletions Grafos/slides/arvores_diametro/main.tex
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Original file line number Diff line number Diff line change
Expand Up @@ -1169,6 +1169,155 @@
\inputsnippet{cpp}{27}{41}{codes/dp.cpp}

\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};

\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};

\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };
\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{Partindo da idéia anterior é possível guardar apenas os dois melhores valores. } };

\end{tikzpicture}
\end{frame}


\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};

\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };
\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{Partindo da idéia anterior é possível guardar apenas os dois melhores valores. } };
\node[anchor=west] (a) at (0.1, 5.0) { $\star$ \bbtext{Guardaremos em $\displaystyle \mathrm{d1}$ e $\displaystyle \mathrm{d2}$, o maior e segundo maior valor para $\displaystyle \mathrm{to\_leaf}[v]$.} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };
\node[anchor=west] (a) at (0.5, 5.0) { $\star$ \bbtext{Caso contrário o maior valor será somado.} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };
\node[anchor=west] (a) at (0.5, 5.0) { $\star$ \bbtext{Caso contrário o maior valor será somado.} };


\node[anchor=west] (a) at (0.1, 4.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2}$ tem-se que :} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };
\node[anchor=west] (a) at (0.5, 5.0) { $\star$ \bbtext{Caso contrário o maior valor será somado.} };


\node[anchor=west] (a) at (0.1, 4.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2}$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 3.5) { $\star$ \bbtext{Se $u$ tiver pelo menos dois filhos somará os dois maiores valores.} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };
\node[anchor=west] (a) at (0.5, 5.0) { $\star$ \bbtext{Caso contrário o maior valor será somado.} };


\node[anchor=west] (a) at (0.1, 4.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2}$ tem-se que :} };

\node[anchor=west] (a) at (0.5, 3.5) { $\star$ \bbtext{Se $u$ tiver pelo menos dois filhos somará os dois maiores valores.} };
\node[anchor=west] (a) at (0.5, 3.0) { $\star$ \bbtext{Se $u$ apenas um filho: $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2} = 2 + d1 + (-1) = 1 + d1$} };

\end{tikzpicture}
\end{frame}
\begin{frame}[plain,t]

\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
\node[draw,opacity=0] at (14, 8) {x};
\node[anchor=west] (title) at (0.0, 7.0) { \Large \bbbold{Possível simplificação da DFS} };

\node[anchor=west] (a) at (0.1, 6.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1$ tem-se que :} };
\node[anchor=west] (a) at (0.5, 5.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{to\_leaf}[u] = max(\displaystyle \mathrm{d1}, \displaystyle \mathrm{d2}) + 1 = -1 + 1= 0$} };
\node[anchor=west] (a) at (0.5, 5.0) { $\star$ \bbtext{Caso contrário o maior valor será somado.} };


\node[anchor=west] (a) at (0.1, 4.0) { $\star$ \bbtext{ Fazendo com que $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2}$ tem-se que :} };

\node[anchor=west] (a) at (0.5, 3.5) { $\star$ \bbtext{Se $u$ tiver pelo menos dois filhos somará os dois maiores.} };
\node[anchor=west] (a) at (0.5, 3.0) { $\star$ \bbtext{Se $u$ apenas um filho: $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2} = 2 + d1 + -1 = 1 + d1$} };
\node[anchor=west] (a) at (0.5, 2.5) { $\star$ \bbtext{Se $u$ for uma folha: $\displaystyle \mathrm{max\_lenght}[u] = 2 + \displaystyle \mathrm{d1} + \displaystyle \mathrm{d2} = 2 + -1 + -1 = 0$} };

\end{tikzpicture}
\end{frame}

\begin{frame}[plain,t]

\inputsnippet{cpp}{10}{30}{codes/dp2.cpp}

\end{frame}

\begin{frame}[plain,t]
\begin{tikzpicture}
\node[draw,opacity=0] at (0, 0) {x};
Expand Down