interrupt for NNDescent (#3432)

Summary:

Addresses the issue in #3173 

for `IndexNNDescent`, I see that there is already interrupt implemented for it's [search](https://fburl.com/code/iwn3tqic) API, so I looked into it's `add` API.

For a given dataset nb = 10 mil, iter = 10, K =  32, d = 32 on a CPU only machine reveals that bulk of the cost comes from [nndescent](https://fburl.com/code/5rdb1p5o). For every iteration of `nndescent` takes around ~12 seconds, ~70-80% of the time is spent on `join` method (~10 seconds per iteration) and ~20-30% spent on `update` (~2 second per iteration). Adding the interrupt on the `join` should suffice on quickly terminating the program when users hit ctrl+C (happy to move the interrupt elsewhere if we think otherwise)

Reviewed By: junjieqi, mdouze

Differential Revision: D57300514

faiss/impl/NNDescent.cpp

@@ -154,15 +154,20 @@ NNDescent::NNDescent(const int d, const int K) : K(K), d(d) {
 NNDescent::~NNDescent() {}
 
 void NNDescent::join(DistanceComputer& qdis) {
+    idx_t check_period = InterruptCallback::get_period_hint(d * search_L);
+    for (idx_t i0 = 0; i0 < (idx_t)ntotal; i0 += check_period) {
+        idx_t i1 = std::min(i0 + check_period, (idx_t)ntotal);
 #pragma omp parallel for default(shared) schedule(dynamic, 100)
-    for (int n = 0; n < ntotal; n++) {
-        graph[n].join([&](int i, int j) {
-            if (i != j) {
-                float dist = qdis.symmetric_dis(i, j);
-                graph[i].insert(j, dist);
-                graph[j].insert(i, dist);
-            }
-        });
+        for (idx_t n = i0; n < i1; n++) {
+            graph[n].join([&](int i, int j) {
+                if (i != j) {
+                    float dist = qdis.symmetric_dis(i, j);
+                    graph[i].insert(j, dist);
+                    graph[j].insert(i, dist);
+                }
+            });
+        }
+        InterruptCallback::check();
     }
 }