e-olymp 2267. Journey

The army of Rzeczpospolita is moving from the city Kostroma to the village Domnino. Two hetmans, Stefan and Konstantin, lead the army.

Stefan procured the roadmap of Kostroma province, and every night he routes the army from one village to the other along some road. Konstantin bought the map of secret trails between villages in advance, and every day he leads the march along the one of such trails. Each hetman asks their guide Ivan Susanin for a route before each march.

The length of each road is indicated on Stefan’s map. So Stefan knows the minimal distance from each village to the Domnino village according to his map. Similarly Konstantin knows the minimal distance from each village to Domnino village along trails on his map.

Ivan Susanin does not want to be disclosed as a secret agent, so each time he chooses a road (for Stefan) or a trail (for Konstantin) so, that the minimal distance to the Domnino village according to the map owned by the asking hetman is strictly decreasing.
Help Ivan to find the longest possible route to the Domnino village.


The first line contains three integer numbers [latex]n, s[/latex] and [latex]t[/latex] — number of villages in Kostroma province, and numbers of start and Domnino village [latex](2 \le n \le 1000; 1 \le s; t \le n)[/latex]. Villages are numbered from [latex]1[/latex] to [latex]n[/latex]. Start and Domnino villages are distinct.

Two blocks follow, the first one describing Stefan’s map, and the second one describing Konstantin’s map.

The first line of each block contains an integer number [latex]m[/latex] — the number of roads/trails between villages [latex](n-1 \le m \le 100000)[/latex]. Each of the following [latex]m[/latex] lines contains three integer numbers [latex]a, b[/latex], and [latex]l[/latex] — describing the road/trail between villages [latex]a[/latex] and [latex]b[/latex] of length [latex]l[/latex] [latex](1 \le a, b \le n; 1 \le l \le 10^6)[/latex].

Rzeczpospolita army can move in any direction along a road or a trail. It’s guaranteed that one can travel from any village to any other using each of the maps. The army starts its movement in the evening from the village number and moves one road each night and one trail each day.


Output the total length of the longest route that Ivan Susanin can arrange for Rzeczpospolita army before reaching the Domnino village (along the roads and trails). If Ivan Susanin can route the army forever without reaching the Domnino village, output the number «[latex]-1[/latex]».


Input Output
1 5 1 5
1 2 2
1 4 2
2 3 1
3 4 1
5 3 1
1 2 2
2 4 2
2 3 1
2 5 2
2 3 1 3
1 2 10
2 3 10
1 3 20
2 3 30
2 1 10
1 3 10
1 1 10
2 3 10


The problem has been resolved together with Sploshnov Kirill.

So, we are dealing with a rather cumbersome task for the graphs, therefore we analyze it consistently. To get started we define the data structure

because dealing with the routes and subsequently, we will have to color our edges. For convenience, we don’t think about two maps as about different graphs, and can establish one graph, where edges of each map are painted in a different color.
For example edges of first map color in RED, and the other in BLUE. Then select the first map is equivalent to passing by red edges, or blue otherwise. In this way, route, that we are looking for, should be based on the successively alternating colors of the edges.
Proceed directly to the solution.
From the condition is understandable, that each hetman knows the shortest path to the final village. This data will be needed for us too, so for each map (edges of the same color) use Dijkstra’s algorithm and find the shortest path from each vertex to the target.  (Function   void djikstra(vector<Route>* graph, int* distancesInColoredGraph, unsigned int quantityOfAllVertices, int finishVertex, int color); ).  We need absolutely standard Dijkstra’s algorithm with the only difference that the edges of the opposite color aren’t available. You can learn more about Dijkstra’s algorithm in the sources of information listed at the end of the article.
Continue analyzing the condition, we understand, that we can’t pass over the edges so, that the shortest distance to the final vertex increased. This will help us to simplify the graph, and significantly reduce the number of possible variants of passage, namely, any bidirectional edge will be either removed completely or strictly directed.  Then, passing on to the edges of the same color in each map, if it doesn’t satisfy the specified condition coloring it as DELETED. (Function  void simplify(vector<Route>* graph, int* distance, unsigned int quantityOfAllVertices, int color); ).
Now we can get started with the search for the longest route. There are two options: either there is the longest path, or we can walk along the edges infinitely, if it does not contradict the statement of the problem, that is, in the combined of two maps graph there is a cycle. So we organize checks on acyclic. Now we have the right to pass along the edges only alternating colors at each step. In order to find a cycle, we use vertex coloring, and will explore the graph until we try to treat already colored vertex or not conclude that it is acyclic.  (Function  bool cyclicDFS(vector<Route>* graph, int* passedInRedGraph, int* passedInBlueGraph, int currentVertex, int color); ). This algorithm can be obtained after detailed acquaintance with the usual cycle searching algorithm (reference to the source is located at the end of the article). If we find any loop in this graph, then our job is over and we should just output «[latex]-1[/latex]».
Otherwise, make sure that the graph is acyclic, we are looking for the longest route. As our graph has been simplified and has no cycles, and all edges are directed, then the task of finding this way becomes computationally simple. For this declaring an array of longest distance dynamically, along the way memorizing the already calculated values, sequentially find the maximum length of the route until we arrive at the finish village. (Function  int maxDistDFS(vector<Route>* graph, int* maxDistancesInRedGraph, int* maxDistancesInBlueGraph, int currentVertex, int color) ). This will be the answer to the task.

Rest details of the implementation can be found in the code of the program or in the sources of information listed at the end of the article.



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