Timus : 1002. Phone Numbers 题解
把电话号码转换成为词典中能够记忆的的单词的组合,找到最短的组合。
我这道题应用到的知识点:
1 Trie数据结构
2 map的应用
3 动态规划法Word Break的知识
4 递归剪枝法
思路:
1 建立Trie字典树。方便查找, 可是字典树不是使用字符来建立的。而是把字符转换成数字。建立一个数字字典树。 然后叶子节点设置一个容器vector<string>装原单词。
2 动态规划建立一个表,记录能够在字典树中找到的字符串的前缀子串
3 假设找到整个串都在字典树中,那么就能够直接返回这个单词。假设无法直接找到。那么就要在表中找到一个前缀子串,然后后面部分在字典树中查找,看是否找到包括这个子串的单词,而且要求找到的单词长度最短。- 这里能够使用剪枝法提高效率。
原题:http://acm.timus.ru/problem.aspx?space=1&num=1002
作者:靖心 - http://blog.csdn.net/kenden23
#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <algorithm>
#include <unordered_map> using namespace std; class PhoneNumber1002_2
{
static const int SIZE = 10;
struct Node
{
vector<string> words;
Node *children[SIZE];
explicit Node () : words()
{
for (int i = 0; i < SIZE; i++)
{
children[i] = NULL;
}
}
}; struct Trie
{
Node *emRoot;
int count;
explicit Trie(int c = 0) : count(c)
{
emRoot = new Node;
}
~Trie()
{
deleteTrie(emRoot);
}
void deleteTrie(Node *root)
{
if (root)
{
for (int i = 0; i < SIZE; i++)
{
deleteTrie(root->children[i]);
}
delete root;
root = NULL;
}
}
}; void insert(Trie *trie, string &keys, string &keyWords)
{
int len = (int)keys.size(); Node *pCrawl = trie->emRoot;
trie->count++; for (int i = 0; i < len; i++)
{
int k = keys[i] - '0';
if (!pCrawl->children[k])
{
pCrawl->children[k] = new Node;
}
pCrawl = pCrawl->children[k];
}
pCrawl->words.push_back(keyWords);
} Node *search(Node *root, string &keys)
{
int len = (int)keys.size(); Node *pCrawl = root;
for (int i = 0; i < len; i++)
{
int k = keys[i] - '0';
if (!pCrawl->children[k])
{
return NULL;//没走全然部keys
}
pCrawl = pCrawl->children[k];
}
return pCrawl;
} void searchLeft(Node *leaf, Node *r, int len, int &prun)
{
if (len >= prun) return; if (leaf->words.size())
{
r = leaf;
prun = len;
return;
} for (int i = 0; i < SIZE; i++)
{
searchLeft(leaf->children[i], r, len+1, prun);
}
} void wordsToKey(string &keys, string &keyWords,
unordered_map<char, char> &umCC)
{
for (int i = 0; i < (int)keyWords.size(); i++)
{
keys.push_back(umCC[keyWords[i]]);
}
} void charsToMap(const string phdig[], unordered_map<char, char> &umCC)
{
for (int i = 0; i < 10; i++)
{
for (int k = 0; k < (int)phdig[i].size(); k++)
{
umCC[phdig[i][k]] = i + '0';
}
}
} string searchComb(Trie *trie, string &num)
{
vector<string> tbl(num.size());
for (int i = 0; i < (int)num.size(); i++)
{
string s = num.substr(0, i+1);
Node *n = search(trie->emRoot, s);
if (n && n->words.size())
{
tbl[i].append(n->words[0]);
continue;//这里错误写成break! 。
}
for (int j = 1; j <= i; j++)
{
if (tbl[j-1].size())
{
s = num.substr(j, i-j+1);
n = search(trie->emRoot, s);
if (n && n->words.size())
{
tbl[i].append(tbl[j-1]);
tbl[i].append(" ");
tbl[i].append(n->words[0]);
break;
}
}
}
} if (tbl.back().size())
{
return tbl.back();
} string ans;
for (int i = 0; i < (int)tbl.size() - 1; i++)
{
if (tbl[i].size())
{
string tmp = tbl[i];
string keys = num.substr(i+1);
Node *n = search(trie->emRoot, keys); if (!n) continue; Node *r = NULL;
int prun = INT_MAX;
searchLeft(n, r, 0, prun); tmp += r->words[0]; if (ans.empty() || tmp.size() < ans.size())
{
ans = tmp;
}
}
}
return ans.empty()? "No solution." : ans;
} //測试函数。不使用解题
void printTrie(Node *n)
{
if (n)
{
for (int i = 0; i < SIZE; i++)
{
printTrie(n->children[i]);
for (int j = 0; j < (int)n->words.size(); j++)
{
cout<<n->words[j]<<endl;
}
}
}
}
public:
PhoneNumber1002_2()
{
const string phdig[10] =
{"oqz","ij","abc","def","gh","kl","mn","prs","tuv","wxy"};
unordered_map<char, char> umCC;
charsToMap(phdig, umCC); int N; string num, keys, keyWords;
while ((cin>>num) && "-1" != num)
{
cin>>N; Trie trie;
while (N--)
{
cin>>keyWords;
wordsToKey(keys, keyWords, umCC); insert(&trie, keys, keyWords); keys.clear();//别忘记清空
} cout<<searchComb(&trie, num)<<endl;
}
}
};
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