这是 meelo 原创的 IEEEXtreme极限编程大赛题解
Xtreme 10.0 - Mancala'h
题目来源 第10届IEEE极限编程大赛
https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/mancalah
You and your friend Alessandro are taking part to an archaeological mission that aims to explore a newly discovered tomb of an ancient pharaoh in Egypt. After an adventurous trip through tunnels, doors and rooms, you and your fellow archaeologists arrive in front of a huge closed door and find a mysterious artefact that appears to be a sort of puzzle for opening the door. The artefact is composed of some inscriptions and a massive wooden disc with many bins carved at the perimeter. Some of the bins contain seeds.
On a side of the artefact, an inscription says: "Harvest all the seeds in the first bin. Sow them in the following bins, one-by-one. Rotate the disc by one step. Repeat."
On the other side of the artefact, a second inscription says: "How far you can go before starting to endless repeat the same harvesting pattern? How far can you go backwards?"
After a fellow archaeologist translated the inscriptions, your friend Alessandro exclaims: "I know this game: it’s Mancala’h! I know that from every possible configuration, the game evolves to a periodic status, meaning that at some point you start repeating the same pattern. For example, the configuration [1, 1, 1] evolves to [0, 2, 1] because you take the seed in the first bin and put it in the second bin. Then, after a clockwise rotation of the disc, the second bin with two seeds becomes the first and we can remove the leading zero from notation. So, the configuration is now [2, 1], which evolves to [2, 1] itself, because you take the two seeds in the first bin and put one in the second bin and one in the third bin, which was empty. In this case we can say that the depth of the original [1, 1, 1] configuration is equal to 1, meaning that in a single step we reach a periodic status."
"The situation can be more complex, for example the configuration [4] evolves to [1, 1, 1, 1], which in turn evolves to [2, 1, 1], which evolves to [2, 2], which evolves to [3, 1], which evolves again to [2, 1, 1], and so on. In this case the depth of the configuration [4] is 2 and the period is 3 steps long."
"The first thing we need to find is exactly the depth of the configuration we have found here."
"Going backwards in the game evolution is not as easy as going forward: for example, we have seen that configurations [1, 1, 1, 1] and [3, 1] evolve both to [2, 1, 1] in one step. On the contrary, the configuration [1, 2] cannot be the evolution of any valid configuration, because we exclude all the so-called not connectedconfigurations, which are those that contain a zero, such as [1, 0, 2]."
"The second thing we need to find, in fact, is the height of the given configuration, which is equal to the maximum number of backwards steps or, in other words, the distance (in terms of number of steps) of the farthest previous configuration. We cannot iterate backwards over the period, so the height is the length of the longest sequence of unique configurations that leads to the current configuration."
Help the archaeologists to solve the mystery by finding the depth D and the height H of given Mancala’h configurations.
Input Format
The input contains a single Mancala’h configuration. A Mancala’h configuration is defined by the sequence [N1, N2, N3, ..., NL] of L (1 ≤ L ≤ 100) integers separated by a blank-space. Each integer Ni (1 ≤ Ni ≤ 500) represents the number of seeds in the ith bin.
Constraints
The size of the Mancala'h board is large enough that you will never have a board in which all of the bins are filled. In other words, there are always more bins than seeds.
The number of unique configurations reachable from any input (either forwards or backwards) is at most 5 * 106.
Output Format
The output is a single line containing two integers D and H, separated by a blank-space. The first integer D is the depth of the Mancala’h configuration specified in the input. The second integer H is the height of the Mancala’h configuration specified in the input.
Sample Input
2 1 1
Sample Output
0 3
Explanation
The depth of the Mancala’h configuration [2, 1, 1] is 0 because the period includes the given configuration:
[2, 1, 1] → [2, 2] → [3, 1] → [2, 1, 1]
The height is 3 because there are three backwards evolutions, two of which contain 3 steps:
[2, 1, 1] ← [1, 1, 1, 1] ← [4] ← [1, 3]
[2, 1, 1] ← [3, 1] ← [2, 2] ← [1, 1, 2]
[2, 1, 1] ← [3, 1] ← [1, 2, 1]
Note that for the following is not a valid backwards evolution because it repeats the configuration [2, 1, 1]:
[2, 1, 1] ← [3, 1] ← [2, 2] ← [2, 1, 1]
As an additional example, suppose that the input was:
3 1 2 3 2
For this Mancala’h, the depth is 6:
[3, 1, 2, 3, 2] → [2, 3, 4, 2] →[4, 5, 2] →
[6, 3, 1, 1] → [4, 2, 2, 1, 1, 1] → [3, 3, 2, 2, 1] →
[4, 3, 3, 1] → [4, 4, 2, 1] → [5, 3, 2, 1] →
[4, 3, 2, 1, 1] → [4, 3, 2, 2] → [4, 3, 3, 1]
The height is 1 because the only previous configuration possible is:
[3, 1, 2, 3, 2] ← [1, 2, 1, 2, 3, 2]题目解析
状态从前往后推只有一种可能性，但是可能存在循环，如何判断是否存在循环是一个关键的问题。事实上，只需要保存之前访问过的状态，线性遍历一遍，并不需要在常数时间内判断是否位于循环节内。
状态从后往前推可能有多种可能性，同样可能存在循环，判断是否存在循环的方法与从前往后推相同。存在多种可能性，那么状态会构成一棵树，需要进行深度优先搜索。
另一个关键问题是如何存储一个状态。状态从前往后推时，会去除第一个元素，需要访问中间的元素有可能在后面添加元素；状态从后往前推时，会添加一个元素，需要访问中间的元素，有可能需要删除末位的元素。
双端队列deque能够很好的满足这些要求，能够在常数时间在首尾添加或者删除元素，同时能在常数时间访问每一个元素。
程序
C++
#include <vector> #include <iostream> #include <algorithm> #include <deque> using namespace std; // return: >=0 found, -1 not found int find(const vector<deque<int> > &states, const deque<int> &state) { for(int i=0; i<states.size(); i++) { if(states[i] == state) { return i; } } return -1; } int depth(deque<int> s) { int index = 0; vector<deque<int> > states; while(find(states, s) == -1) { states.push_back(s); // get next state int front = s.front(); s.pop_front(); int i, q_size = s.size(); for(i=0; i<min(front, q_size); i++) { s[i]++; } for(;i<front; i++) { s.push_back(1); } } return find(states, s); } // depth first search int height(deque<int> &state, vector<deque<int> > &statesBefore) { int max_depth = 0; for(int l=state.size()-1; l>=0; l--) { deque<int> s(state); for(int i=l; i>=0; i--) { s[i]--; } bool bad_state = false; // true: contain a zero for(int i=0; i<s.size()-1; i++) { if(s[i]==0 && s[i+1]!=0) { bad_state = true; break; } } if(!bad_state) { s.push_front(l+1); // remove trailing zero while(s.back() == 0) { s.pop_back(); } if(find(statesBefore, s) != -1 ) break; statesBefore.push_back(s); max_depth = max(max_depth, height(s, statesBefore)+1) ; statesBefore.pop_back(); } } return max_depth; } int main() { deque<int> state; int m; while(cin >> m) { state.push_back(m); } vector<deque<int> > statesBefore; statesBefore.push_back(state); cout << depth(state) << ' ' << height(state, statesBefore); return 0; }