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Diffstat (limited to 'other/count_possible_routes.cpp')
-rw-r--r-- | other/count_possible_routes.cpp | 42 |
1 files changed, 42 insertions, 0 deletions
diff --git a/other/count_possible_routes.cpp b/other/count_possible_routes.cpp new file mode 100644 index 0000000..228fb2e --- /dev/null +++ b/other/count_possible_routes.cpp @@ -0,0 +1,42 @@ +// 1575. Count All Possible Routes +#include "leetcode.h" + +/* + * given an array of disticnt positive integers locations where 'locations[i]' + * represents the position of city 'i'. you are lso given integers 'start, + * finish, fuel' representing the starting city, ending city, and the initial + * amount of fuel you have, respectively. at each step if you are in city 'i', + * you can pick any city 'j' such that 'j != i' and '0 <= j < locations.length' + * and move to city 'j'. moving city 'i' to 'j' reduces the amount of fuel you + * have by |locations[i] - locations[j]|. Please notice that |x| denotes the + * absolute value of x. notice that fuel cannot become negative at any point in + * time. and that are allowed to visit any city more than once (including + * 'start' and 'finish'). return the count of all possible routes from 'start' + * to 'finish'. since the answer may be large, return it modulo 10^9+7. + */ + +class Solution { +public: + int countRoutes(vector<int> &locations, int start, int finish, int fuel) { + int n = locations.size(), mod = 1e9 + 7, ans = 0; + vvd dp(n, vector<int>(fuel + 1)); + dp[start][fuel] = 1; + for (int f = fuel; f >= 0; f--) + for (int i = 0; i < n; i++) + for (int j = 0; j < n; j++) { + int diff = abs(locations[i] - locations[j]); + if (i != j && f >= diff) + dp[j][f - diff] = (dp[j][f - diff] % mod + dp[i][f] % mod) % mod; + } + for (int i = 0; i <= fuel; i++) + ans = (ans % mod + dp[finish][i] % mod) % mod; + return ans; + } +}; + +int main() { + Solution obj; + vector<int> l1 = {2, 3, 6, 8, 4}, l2 = {4, 3, 1}; + printf("%d\n", obj.countRoutes(l1, 1, 3, 5)); // expect: 4 + printf("%d\n", obj.countRoutes(l2, 1, 0, 6)); // expect: 5 +} |