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+// 1976. Number of Ways to Arrive at Destination
+#include "leetcode.h"
+
+/*
+ * you are in a city that consists of 'n' intersections numbered from 0 to 'n -
+ * 1' with bi directional roads between some intersections. the inputs are
+ * generated such that you can reach any intersection from any other
+ * intersection and that there is at most one road between any two
+ * intersections. you are given an integer 'n' and a 2d integer array 'roads'
+ * where 'roads[i] = [u_i, v_i, t_i]' means that there is a road between
+ * intersections 'u' and 'v' that takes 't' minutes to travel. you want to know
+ * in how many ways you can travel from interseciton 0 to intersection 'n - 1'.
+ */
+
+class Solution {
+public:
+ int countPaths(int n, vector<vector<int>> &roads) {
+ vector<vector<pair<long long, long long>>> graph(n);
+ for (auto &r : roads) {
+ long long u = r[0], v = r[1], t = r[2];
+ graph[u].push_back({v, t});
+ graph[v].push_back({u, t});
+ }
+ return dijkstra(graph, n, 0);
+ }
+
+private:
+ int mod = 1e9 + 7;
+ int dijkstra(const vector<vector<pair<long long, long long>>> &graph, int n,
+ int src) {
+ vector<long long> dist(n, LONG_MAX), ways(n);
+ ways[src] = 1;
+ dist[src] = 0;
+ priority_queue<pair<long long, long long>,
+ vector<pair<long long, long long>>, greater<>>
+ pq;
+ pq.push({0, 0});
+ while (!pq.empty()) {
+ auto [d, u] = pq.top();
+ pq.pop();
+ if (d > dist[u])
+ continue;
+ for (auto [v, t] : graph[u]) {
+ if (dist[v] > d + t) {
+ dist[v] = d + t;
+ ways[v] = ways[u];
+ } else if (dist[v] == d + t)
+ ways[v] = (ways[v] + ways[u]) % mod;
+ }
+ }
+ return ways[n - 1];
+ }
+};
+
+int main() {
+ Solution obj;
+ vvd r1 = {{0, 6, 7}, {0, 1, 2}, {1, 2, 3}, {1, 3, 3}, {6, 3, 3},
+ {3, 5, 1}, {6, 5, 1}, {2, 5, 1}, {0, 4, 5}, {4, 6, 2}};
+ vvd r2 = {{1, 0, 10}};
+ printf("%d\n", obj.countPaths(7, r1)); // expect: 4
+ printf("%d\n", obj.countPaths(2, r2)); // expect: 1
+}