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| const numWays = (n, relation, k) => {
const t = n - 1,
g = new Array(n).fill(0).map(() => new Array());
let ans = 0;
for (const [u, v] of relation)
g[u].push(v);
const dfs = (u, steps) => {
if (steps == k) {
if (u == t) ++ans;
return;
}
for (const v of g[u])
dfs(v, steps + 1);
}
dfs(0, 0);
return ans;
};
|