【编程题】2018阿里编程题题一

转化为最短路径搜索：

将D中存储的词看为一条路径：如”like”表示“ilikealibaba”中存在一条从S[1]: “l”直接到S[5]: “a”的路径。

则最少空格数的解就是从S[0]走到S[n]最短路径的解。

测试样例中存在两条路径：

1: S[0]->S[1]->S[5]->S[7]->S[12]

2: S[0]->S[1]->S[5]->S[12]

最短路径使用BFS

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#include <stdio.h>#include <vector>#include <numeric>#include <algorithm>#include <iostream>#include <string>#include <set>#include <regex>#include <queue>using namespace std;//#define debug_void mincut(const string& str, const set<string>& dict){    vector<vector<int>> path_map(str.size());    vector<int> cost(str.size()+1,99999);    int strSize = str.size();    for (auto i = 0; i < str.size();++i)    {        for (auto iter = dict.begin(); iter != dict.end();iter++)        {            if (str.substr(i,(*iter).size()) == *iter)            {                path_map[i].push_back(i+(*iter).size());            }        }    }    queue<int> que;    que.push(0);    cost[0] = 0;    while (!que.empty())    {        int s = que.front();        que.pop();        for (auto i = 0; i < path_map[s].size();++i)        {            if (cost[s]+1<cost[path_map[s][i]])            {                if (path_map[s][i] != strSize)//不是终点                {                    que.push(path_map[s][i]);                }                cost[path_map[s][i]] = cost[s] + 1;            }        }    }    int cur_index(strSize);    vector<string> print_vec;    if (cost[strSize] == 99999)    {        cout <<"n/a";        return;    }    for (auto i = strSize; i >= 0;--i)    {        if (cost[i] + 1 == cost[cur_index])        {            string tmp = str.substr(i, cur_index-i);            print_vec.push_back(tmp);            cur_index = i;        }    }    for (auto iter = print_vec.rbegin(); iter != print_vec.rend();iter++)    {        cout << *iter << " ";    }}int main(int argc, const char * argv[]){    string strS;    string dictStr;    int nDict;    set<string> dict;#ifdef debug_    strS = "ilikealibaba";    dict.insert("i");    dict.insert("like");    dict.insert("ali");    dict.insert("liba");    dict.insert("baba");    dict.insert("alibaba");#else    cin >> strS;    cin >> nDict;    for (int i = 0; i < nDict; i++)    {        cin >> dictStr;        dict.insert(dictStr);    }#endif    mincut(strS, dict);    return 0;}

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