Fair
Contest: Codeforces Round 485 (Div. 1)
Difficulty rating: 1600
Tags: graphs, greedy, number theory, shortest paths

Some company is going to hold a fair in Byteland. There are $$$n$$$ towns in Byteland and $$$m$$$ two-way roads between towns. Of course, you can reach any town from any other town using roads.

There are $$$k$$$ types of goods produced in Byteland and every town produces only one type. To hold a fair you have to bring at least $$$s$$$ different types of goods. It costs $$$d(u,v)$$$ coins to bring goods from town $$$u$$$ to town $$$v$$$ where $$$d(u,v)$$$ is the length of the shortest path from $$$u$$$ to $$$v$$$. Length of a path is the number of roads in this path.

The organizers will cover all travel expenses but they can choose the towns to bring goods from. Now they want to calculate minimum expenses to hold a fair in each of $$$n$$$ towns.

Input format:
There are $$$4$$$ integers $$$n$$$, $$$m$$$, $$$k$$$, $$$s$$$ in the first line of input ($$$1 \le n \le 10^{5}$$$, $$$0 \le m \le 10^{5}$$$, $$$1 \le s \le k \le min(n, 100)$$$) — the number of towns, the number of roads, the number of different types of goods, the number of different types of goods necessary to hold a fair.

In the next line there are $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_{i} \le k$$$), where $$$a_i$$$ is the type of goods produced in the $$$i$$$-th town. It is guaranteed that all integers between $$$1$$$ and $$$k$$$ occur at least once among integers $$$a_{i}$$$.

In the next $$$m$$$ lines roads are described. Each road is described by two integers $$$u$$$ $$$v$$$ ($$$1 \le u, v \le n$$$, $$$u \ne v$$$) — the towns connected by this road. It is guaranteed that there is no more than one road between every two towns. It is guaranteed that you can go from any town to any other town via roads.

Output format:
Print $$$n$$$ numbers, the $$$i$$$-th of them is the minimum number of coins you need to spend on travel expenses to hold a fair in town $$$i$$$. Separate numbers with spaces.

Constraints:
Time limit: 2.0 s
Memory limit: 512.0 MB

Note:
Let's look at the first sample.

To hold a fair in town $$$1$$$ you can bring goods from towns $$$1$$$ ($$$0$$$ coins), $$$2$$$ ($$$1$$$ coin) and $$$4$$$ ($$$1$$$ coin). Total numbers of coins is $$$2$$$.

Town $$$2$$$: Goods from towns $$$2$$$ ($$$0$$$), $$$1$$$ ($$$1$$$), $$$3$$$ ($$$1$$$). Sum equals $$$2$$$.

Town $$$3$$$: Goods from towns $$$3$$$ ($$$0$$$), $$$2$$$ ($$$1$$$), $$$4$$$ ($$$1$$$). Sum equals $$$2$$$.

Town $$$4$$$: Goods from towns $$$4$$$ ($$$0$$$), $$$1$$$ ($$$1$$$), $$$5$$$ ($$$1$$$). Sum equals $$$2$$$.

Town $$$5$$$: Goods from towns $$$5$$$ ($$$0$$$), $$$4$$$ ($$$1$$$), $$$3$$$ ($$$2$$$). Sum equals $$$3$$$.

Example input:
5 5 4 3
1 2 4 3 2
1 2
2 3
3 4
4 1
4 5

Example output:
2 2 2 2 3

Example input:
7 6 3 2
1 2 3 3 2 2 1
1 2
2 3
3 4
2 5
5 6
6 7

Example output:
1 1 1 2 2 1 1

Editorial:
Let's find a cost to bring a good $$$t$$$ in each town. To do this we will run BFS from all towns producing good $$$t$$$ at once. Just add all that towns in queue and run usual BFS. Complexity of BFS is $$$O(n+m)$$$, so total complexity of $$$k$$$ BFSs will be $$$O(k(n+m))$$$.
Now for each town we should choose $$$s$$$ cheapest goods. We can sort them in $$$O(k \log k)$$$, but we can use nth_element instead. It will put the $$$s$$$-th element in sorted order on place $$$s$$$, and all elements smaller will be to the left. Since we are interested only in their sum, we can just sum up first $$$s$$$ elements after calling nth_element.
Another way to achieve $$$O(k(m+n))$$$ complexity is to run all $$$k$$$ BFSs simultaneously, then for each town first $$$s$$$ goods to reach it are the cheapest.
Bonus: solve the problem in $$$O(s(m+n))$$$ time.