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	<title>Sippert, Tiago A. S. &#187; recursao</title>
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		<title>Algoritmo: O imperador Lu Zhin Du</title>
		<link>http://sippert.com.br/algoritmos/algoritmo-o-imperador-lu-zhin-du-cc/</link>
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		<pubDate>Sat, 13 Sep 2008 14:22:01 +0000</pubDate>
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		<description><![CDATA[Algoritmo de Maratona de Programação utilizando recursão: #include&#60;iostream&#62; using namespace std; int m,n,k,mat[100][100],vis[100][100]; bool recursao(int i, int j){ if(j==n-1) return true; vis[i][j]=1; bool parcial=false; if((j&#60;n-1)&#38;&#38;(!vis[i][j+1])&#38;&#38;(abs(mat[i][j+1]-mat[i][j])==k)) parcial=recursao(i,j+1); if((!parcial)&#38;&#38;(i&#62;0)&#38;&#38;(!vis[i-1][j])&#38;&#38;(abs(mat[i-1][j]-mat[i][j])==k)) parcial=recursao(i-1,j); if((!parcial)&#38;&#38;(j&#62;0)&#38;&#38;(!vis[i][j-1])&#38;&#38;(abs(mat[i][j-1]-mat[i][j])==k)) parcial=recursao(i,j-1); if((!parcial)&#38;&#38;(i&#60;m-1)&#38;&#38;(!vis[i+1][j])&#38;&#38;(abs(mat[i+1][j]-mat[i][j])==k)) parcial=recursao(i+1,j); return parcial; } int main(){ int t=1; while((cin &#62;&#62; m &#62;&#62; n &#62;&#62; k)&#38;&#38;(m!=0)&#38;&#38;(n!=0)){ memset(vis,0,sizeof(vis)); for(int i=0;i&#60;m;i++) for(int j=0;j&#60;n;j++) cin &#62;&#62; mat[i][j]; bool [...]]]></description>
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