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PO
 

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  1. PO
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    <p>I tried this, its fast but I am not sure whether it is correct. When you reach a black node it means that you have already calculated path from that node so I stored the number of paths in the structure of node and used it.</p> <hr> <pre><code>#include &lt;stdio.h&gt; #include &lt;malloc.h&gt; #include &lt;vector&gt; #include &lt;algorithm&gt; #define WHITE 0 #define GRAY 1 #define BLACK 2 using namespace std; typedef struct vertexstruct{ int color; vector&lt;int&gt; edgelist; long paths; }vertex; int cyclenode; int throughend; long dfs(int v,vertex *vertices,int numvertices){ long paths = 0; if(vertices[v].color == BLACK) return vertices[v].paths; if(vertices[v].color == GRAY) { if(cyclenode == 0)cyclenode = v; } if(vertices[v].color == WHITE){ vertices[v].color = GRAY; for(int i=0;i&lt;vertices[v].edgelist.size();i++){ long x = dfs(vertices[v].edgelist[i],vertices,numvertices); if(cyclenode) vertices[v].cycle=1; if(x == -1) return -1; paths += x; } if(cyclenode &amp;&amp; paths &gt;0) return -1; if(v == numvertices){ if(cyclenode) { return -1; } throughend = 0; vertices[v].paths = 1; return 1; } if(cyclenode ==v &amp;&amp; throughend == 0) cyclenode = 0; vertices[v].color = BLACK; vertices[v].paths = paths % 1000000000; } return paths % 1000000000; } void numpaths(int numvertices,int numedges,vertex *vertices){ long numpaths = 0; cyclenode = 0; throughend =0; numpaths = dfs(1,vertices,numvertices); if(numpaths == -1) printf("INFINITE PATHS\n"); else printf("%ld\n",numpaths); } int main(){ int edgecount=0; int numverts,numedges; fscanf(stdin,"%d %d",&amp;numverts,&amp;numedges); vertex verts[numverts+1]; for(int i =1;i&lt;=numverts;i++){ verts[i].color = WHITE; verts[i].paths = 0; } edgecount = 0; int a,b; while(edgecount &lt; numedges){ scanf("%d %d",&amp;a,&amp;b); verts[a].edgelist.push_back(b); edgecount++; } numpaths(numverts,numedges,verts); } </code></pre>
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    1. PONumber of paths between two nodes in a Directed Cyclic Graph
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    1. PTAnswer
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    1. USvgeta
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    1. COThere is a bug in the cycle detection logic. Only one cycle can be detected at a time, and if a path has two cycles the algorithm may break. Consider the input `5 6 1 2 2 1 1 3 3 4 4 3 1 5`. Expected output is `INFINITE PATHS`, but your program outputs `1`.
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    2. CO@tom, in your case. the answer is 1 because there is only one path
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    3. COi mean there is only one path between 1 and 6 which is 1->5->6. the cycle in 2,3 and 3,4 does not matter. what do you think?
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