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# CHEFCCYL - Editorial

Ah, I could have used the [hid e][/h ide] function if I knew it earlier. This would make editorials clearer, I think.

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Contest

Author: Full name

Tester: Full name

Editorialist: Full name

# DIFFICULTY:

CAKEWALK, SIMPLE, EASY, MEDIUM or HARD. Intermediate levels like SIMPLE-EASY, EASY-MEDIUM, MEDIUM-HARD are also possible.

# PREREQUISITES:

Greedy, DP, Math etc. Ideally with the links to Wikipedia

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# QUICK EXPLANATION:

Let's call the endpoints of additional edges "keypoint"s. The topology of keypoints form a cycle, so we can handle distance queries of keypoints quickly. For a general query, let's say it's from vertex $v_1$ on cycle $c_1$ to vertex $v_2$ on cycle $c_2$. The path goes through some keypoint $k_1$ on cycle $c_1$ and some $k_2$ on cycle $c_2$. Since every cycle has at most $2$ keypoints, we enumerate $k_1$ and $k_2$, and thus answer queries on constant time.

# EXPLANATION:

This is a shortest path problem on a graph with $M=A_1+A_2+\dots+A_N$ vertices and $M+1$ edges. We can run Dijkstra's Algorithm for each query. The time complexity for heap-optimized Dijkstra's Algorithm is $O(M\log M)$. So total complexity is $O(QM\log M)$.

### Querying on one cycle

Consider the easier problem: you have one cycle and you want to answer distance queries. Let's number the vertices $1,2,\dots,N$, and suppose the cycle is $1-2-3-\dots-N-1$. For any $i,j(i < j)$, the only simple paths between $i,j$ are:

• $i\to (i+1)\to(i+2)\to\dots\to j$, length $l_1$;
• $i\to (i-1)\to\dots\to 1\to N\to (N-1)\to\dots\to (j+1)\to j$, length $l_2$.

Let's precompute $d_i$ as the length of path $1\to 2\to\dots\to i$. Then $l_1=d_j-d_i$. Since $l_2+l_1$ is the length of the whole cycle, $l_2$ can be easily calculated, too. We return the smaller number between $l_1$ and $l_2$, and answer the query in $O(1)$ time.

### Querying on keypoints

Let's call the endpoints of additional edges "keypoint"s. Consider how to answer distance queries between keypoints. We construct a new graph $C$ whose vertices are only the keypoints. For two keypoints:

• if they are connected by an additional edge, we add this edge to $C$;
• if they are in the same cycle, we query that cycle and get their distance $d$ in the cycle, then connect them with an edge of weight $d$.

The resulting graph $C$ is a cycle: every vertex has degree $2$ and $C$ is connected. Thus we can easily handle distance queries on $C$. Also, $C$ preserves distances on the original graph. So we can answer distance queries about endpoints in $O(1)$ time.

### The final solution

First, using the above algorithms we can support distance queries on one cycle and on keypoints.

Suppose there is a query $(V_1,C_1,V_2,C_2)$. Since $C_1\ne C_2$, their shortest path must pass through some keypoint in cycle $C_1$, suppose it's $k_1$; it also passes through some keypoint in cycle $C_2$, say $k_2$. Then the shortest path has length $dis(V_1,k_1)+dis(k_1,k_2)+dis(k_2,V_2)$, and the smallest this value among all $(k_1,k_2)$'s is the answer. The three $dis()$'s can be obtained in $O(1)$ time. Each cycle has at most $2$ keypoints, so there are at most $4$ pairs of $(k_1,k_2)$ that we need to enumerate.

The time complexity is $O(Q+M)$, where $M=A_1+A_2+\dots+A_N$.

# AUTHOR'S AND TESTER'S SOLUTIONS:

Author's solution can be found [here][333]. Tester's solution can be found [here][444].

# RELATED PROBLEMS:

[333]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server. [444]: The link is provided by admins after the contest ends and the solutions are uploaded on the CodeChef Server

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question asked: 22 Aug '17, 20:25

question was seen: 118 times

last updated: 10 Aug, 16:18