1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
|
// 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
}
|