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  1. PO
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    <p>Let say you need to go from src to dest.</p> <p>With each vertex x associate two values count and val, count is the number of shortest path from src to x and val is the shortest distance from src to x. Also maintain a visited variable telling whether node is arrived first time of not.</p> <p>Apply usual BFS algorithm,</p> <pre><code>Initialize u = src visited[u] = 1,val[u] = count[u] = 1 For each child v of u, if v is not visited </code></pre> <p>A node is visited first time so it has only one path from source till now via u, so shortest path upto v is 1 + shortest path upto u, and number of ways to reach v via shortest path is same as count[u] because say u has 5 ways to reach from source, then only these 5 ways can be extended upto v as v is encountered first time via u, so</p> <pre><code>val[v] = val[u]+1, count[v] = count[u], visited[v] = 1 if v is visited </code></pre> <p>If v is already visited, which mean, there way some other path upto v via some other vertices, then three cases arises:</p> <pre><code>1 :if val[v] == val[u]+1 </code></pre> <p>if current val[v] which is dist upto v via some other path is equal to val[u]+1, i.e we have equal shortest distances for reaching v using current path through u and the other path upto v, then the shortest distance upto v remains same, but the numbr of path increases by number of paths of reaching u.</p> <pre><code>count[v] = count[v]+count[u] 2: val[v] &gt; val[u]+1 </code></pre> <p>If current path of reaching v is smaller than previous value of val[v], then val[v] is stored current path and count[v] is also updated</p> <pre><code>val[v] = val[u]+1 count[v] = count[u] </code></pre> <p>The third case is if current path has distance greater than previous path, in this case, no need to change the values of val[v] and count[v]</p> <p>Do this algorithm till the BFS is complete. In the end val[dest] contain the shortest distance from source and count[dest] contain the number of ways from src to dest.</p>
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    1. USShashwat Kumar
    1. USShashwat Kumar
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    1. COSo, when you are calling the vertex "visited" that means that it is currently in BFS queue? So when it is no longer in the queue I should just skip it whenever I encounter it?
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      1. USMiroslav Lhoťan
    2. COAlso, where do I add to the count value? When I start with zero count in the src vertex there will be zero all the way through.
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      1. USMiroslav Lhoťan
    3. CONo visited means it is visited once. It need not to be in queue at that time. For count, it was my mistake. Initialize count[src] as 1 as number of ways to reach from src to src is one.
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