21483_HeatWave

The good folks in Texas are having a heatwave this summer. Their Texas Longhorn cows make for good eating but are not so adept at creating creamy delicious dairy products. Farmer John is leading the charge to deliver plenty of ice cold nutritious milk to Texas so the Texans will not suffer the heat too much.

FJ has studied the routes that can be used to move milk from Wisconsin to Texas. These routes have a total of T (1 ≤ T ≤ 2,500) towns conveniently numbered 1..T along the way (including the starting and ending towns). Each town (except the source and destination towns) is connected to at least two other towns by bidirectional roads that have some cost of traversal (owing to gasoline consumption, tolls, etc.). Consider this map of seven towns; town 5 is the source of the milk and town 4 is its destination (bracketed integers represent costs to traverse the route):

                             [1]----1---[3]-
                            /               \
                     [3]---6---[4]---3--[3]--4
                    /               /       /|
                   5         --[3]--  --[2]- |
                    \       /        /       |
                     [5]---7---[2]--2---[3]---
                           |       /
                          [1]------

Traversing 5-6-3-4 requires spending 3 (5->6) + 4 (6->3) + 3 (3->4) = 10 total expenses.

Given a map of all the C (1 ≤ C ≤ 6,200) connections (described as two endpoints R_1i and R_2i (1 ≤ R_1i ≤ T; 1 ≤ R_2i ≤ T) and costs (1 ≤ C_i ≤ 1,000), find the smallest total expense to traverse from the starting town T_s (1 ≤ T_s ≤ T) to the destination town T_e (1 ≤ T_e ≤ T).

Line 1:        Four space-separated integers: T, C, T_s, and T_e

Lines 2..C+1:  Line i+1 describes road i with three space-separated integers: R_1i, R_2i, and C_i

Line 1:  A single integer that is the length of the shortest route from T_s to T_e. It is guaranteed that at least one route exists.

7 11 5 4
2 4 2
1 4 3
7 2 2
3 4 3
5 7 5
7 3 3
6 1 1
6 3 4
2 4 3
5 6 3
7 2 1

7

5->6->1->4 (3 + 1 + 3)