HDU 4172解题报告

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Octagons

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 217 Accepted Submission(s): 82

Problem Description
Below is a picture of an infinite hyperbolic tessellation of octagons. If we think of this as a graph of vertices (of degree three), then there exists an isomorphism of the graph which maps any vertex x onto any other vertex y. Every edge is given a label from the set {a,b,c} in such a way that every vertex has all three types of edges incident on it, and the labels alternate around each octagon. Part of this labeling is illustrated in the diagram.

So a path in this graph (starting from any vertex) can be specified by a sequence of edge labels. Your job is to write a program which, given a squence of labels such as “abcbcbcabcaccabb”, returns “closed” if the path ends on the same vertex where it starts, and returns “open” otherwise.

Input
The input will begin with a number Z ≤ 200 on a line by itself. This is followed by Z lines, each of which is a squence of length at least 1 and at most 40 of ‘a’s ‘b’s and ‘c’s.

Output
For each input instance, the output will be the words “closed” or “open”, each on a single line.

Sample Input
2
abababab
abcbcbcbcba

Sample Output
closed
open

这道题题意就是有 一堆 八边形 每条边可以用a b c中任一个标记
相邻两条边的标记不同
每个八边形中只有两个标记
比如： babababa
或 cacacaca

输入一个字串
问能不能按字串的顺序来走八边形 最终回到起点

这道题其实就是字符串处理：
根据几何知识可以知道
规则1 abababa 可换成 b 八边形两个方向走 走7条边和走另一方向的1条边是一样的
规则2 ababab 可换成 ba 6条边 换 两条边
规则3 ababa 可换成 bab 5条边换3条边
规则4 bb 可去掉 一条边来回走

然后仔细点可以发现
只需要 规则3 和 规则4 就可以实现所有规则

不说了
上代码

/*author: Ticholas-Huang*//*2016.4.3*/#include <iostream>#include <string>using namespace std;bool transfer(string ss){    size_t size = ss.size();    if (size == 0)        return true;    for (int i = 0; i + 1 < size; i++)    {        if (ss[i] == ss[i + 1])            return transfer(ss.substr(0, i) + ss.substr(i + 2));    }    for (int i = 0; i + 4 < size; i++)    {        if (ss[i] == ss[i + 2] && ss[i] == ss[i + 4] && ss[i + 1] == ss[i + 3])            return transfer(ss.substr(0, i) + ss.substr(i + 1,3) + ss.substr(i + 5));    }    return false;}int main(int argc, char const *argv[]){    int Case = 0;    cin >> Case;    string testStr;    while (Case)    {        cin >> testStr;        if (transfer(testStr))        {            cout << "closed" << endl;        }        else        {            cout << "open" << endl;        }        Case--;    }    return 0;}