算法问题实战策略 WORDCHAIN

地址  https://algospot.com/judge/problem/read/WORDCHAIN

解答:

1 书上的解法是制作有向图 然后查找欧拉回路  代码实现稍后

假设一定存在欧拉路径的做法

 #include <iostream>
 #include <vector>
 #include <string>
 #include <algorithm>

 using namespace std;

 vector<vector<int>> adj;

 vector<string> graph[][];

 vector<int> indegree, outdegree;

 void makeGraph(const vector<string>& words)
 {
     for (int i = ; i < ; i++)
         for (int j = ; j < ; j++)
             graph[i][j].clear();
     adj = vector<vector<int>>(, vector<int>(, ));
     indegree = outdegree = vector<int>(, );
     for (int i = ; i < words.size(); i++) {
         int a = words[i][] - 'a';
         int b = words[i][words[i].size() - ] - 'a';
         graph[a][b].push_back(words[i]);
         adj[a][b]++;
         outdegree[a]++;
         indegree[b]++;
     }
 }

 void getEulerCircuit( int here,vector<int>& circuit )
 {
     for (int there = ; there < adj.size(); ++there) {
         while (adj[here][there] > ) {
             adj[here][there]--;
             getEulerCircuit(there, circuit);
         }
     }

     circuit.push_back(here);
 }

 vector<int> getEulerTrailOrCircuit()
 {
     vector<int> circuit;

     for (int i = ; i < ; i++) {
         if (outdegree[i] == indegree[i] + ) {
             getEulerCircuit(i, circuit);
             return circuit;
         }
     }

     for (int i = ; i < ; ++i)
     {
         if (outdegree[i]) {
             getEulerCircuit(i, circuit);
             return circuit;
         }
     }

     return circuit;
 }

 string solve(const vector<string>& words)
 {
     makeGraph(words);

     vector<int> circuit = getEulerTrailOrCircuit();

     if (circuit.size() != words.size() + ) return "IMPOSSIBLE";

     reverse(circuit.begin(), circuit.end());
     string ret;
     for (int i = ; i < circuit.size(); i++) {
         int a = circuit[i - ], b = circuit[i];
         if (ret.size()) ret += " ";
         ret += graph[a][b].back();
         graph[a][b].pop_back();
     }

     return ret;
 }

 int n, m;
 int main()
 {
     cin >> n;
     while (n--) {
         cin >> m;
         vector<string> words;
         while (m--) {
             string s;
             cin >> s;
             words.push_back(s);
         }

         cout << solve(words) << endl;

     }

     return ;
 }

先统计有向图的出入度 判断有无欧拉回路与欧拉路径后 在进行dfs的做法

 #include <iostream>
 #include <vector>
 #include <string>
 #include <algorithm>

 using namespace std;

 vector<vector<int>> adj;

 vector<string> graph[][];

 vector<int> indegree, outdegree;

 void makeGraph(const vector<string>& words)
 {
     for (int i = ; i < ; i++)
         for (int j = ; j < ; j++)
             graph[i][j].clear();
     adj = vector<vector<int>>(, vector<int>(, ));
     indegree = outdegree = vector<int>(, );
     for (int i = ; i < words.size(); i++) {
         int a = words[i][] - 'a';
         int b = words[i][words[i].size() - ] - 'a';
         graph[a][b].push_back(words[i]);
         adj[a][b]++;
         outdegree[a]++;
         indegree[b]++;
     }
 }

 void getEulerCircuit(int here, vector<int>& circuit)
 {
     for (int there = ; there < adj.size(); ++there) {
         while (adj[here][there] > ) {
             adj[here][there]--;
             getEulerCircuit(there, circuit);
         }
     }

     circuit.push_back(here);
 }

 string solve(const vector<string>& words)
 {
     makeGraph(words);

     int start = -; int end = -;

     for (int i = ; i < ; i++) {
         if (outdegree[i] != indegree[i]) {
             if (outdegree[i] == indegree[i] +  ) {
                 if (- == start)
                     start = i;
                 else
                     return "IMPOSSIBLE";
             }
             else if (outdegree[i] +  == indegree[i] ) {
                 if( - == end)
                     end = i;
                 else
                     return "IMPOSSIBLE";
             }
         }
     }

     vector<int> circuit;
     if (start == - && end == -) {
         for (int i = ; i < ; ++i)
         {
             if (outdegree[i]) {
                 getEulerCircuit(i, circuit);
                 break;
             }
         }
     }
     else {
         getEulerCircuit(start, circuit);
     }

     reverse(circuit.begin(), circuit.end());
     string ret;
     for (int i = ; i < circuit.size(); i++) {
         int a = circuit[i - ], b = circuit[i];
         if (ret.size()) ret += " ";
         ret += graph[a][b].back();
         graph[a][b].pop_back();
     }

     return ret;
 }

 int n, m;
 int main()
 {
     cin >> n;
     while (n--) {
         cin >> m;
         vector<string> words;
         while (m--) {
             string s;
             cin >> s;
             words.push_back(s);
         }

         cout << solve(words) << endl;

     }

     return ;
 }

2 个人觉得 可以使用首尾单词作为关键字 去哈希 然后进行哈希查找与DFS结合的搜索 看看最后能否将单词全部使用 代码如下

TLE

 #include <iostream>
 #include <vector>
 #include <string>

 using namespace std;

 int n, m;

 /*
 3
 4
 dog
 god
 dragon
 need
 3
 aa
 ab
 bb
 2
 ab
 cd

 need dog god dragon
 aa ab bb
 IMPOSSIBLE
 */

 vector<string> ret;

 void Dfs( int begIdx,vector<vector<vector<string>>>& vvstr)
 {
     if (ret.size() == m) {
         return;
     }

     for (int i = ; i < ; i++) {
         for (int j = ; j < vvstr[begIdx][i].size(); j++) {
             //选择 nextBegIdx 开头的字母
             //如果已经选择过了 则进行下一次尝试
             if (vvstr[begIdx][i][j][] == '#')
                 continue;

             //开始选择该单词接龙
             int nextBegIdx = vvstr[begIdx][i][j].back() - 'a';
             ret.push_back(vvstr[begIdx][i][j]);
             vvstr[begIdx][i][j][] = '#'; //修改字符串的第一个字符 作为已经被选择的标记

             Dfs( nextBegIdx, vvstr);

             if (ret.size() == m) {
                 return;
             }

             vvstr[begIdx][i][j][] = 'a'+ begIdx;
             ret.pop_back();
             break;
         }
     }

 }

 void DfsStart(vector<vector<vector<string>>>& vvstr)
 {
     for (int i = ; i < ; i++) {
         for (int j = ; j < ; j++) {
             for (int k = ; k < vvstr[i][j].size(); k++) {
                 //选中这个单词作为开始
                 int nextBegIdx = vvstr[i][j][k].back() - 'a';
                 ret.push_back(vvstr[i][j][k]);
                 vvstr[i][j][k][] = '#'; //修改字符串的第一个字符 作为已经被选择的标记
                 Dfs( nextBegIdx,vvstr);

                 if (ret.size() == m) {
                     return;
                 }

                 //复原 继续下一次选择
                 vvstr[i][j][k][] = 'a'+i;
                 ret.pop_back();
                 break;
             }
         }
     }

 }

 int main()
 {
     cin >> n;
     while (n--) {
         cin >> m;
         string s; vector<vector<vector<string>>> vvstr(, vector<vector<string>>(, vector<string>()));
         ret.clear();
         for (int i = ; i < m; i++){
             cin >> s;
             int beg = s[] - 'a';
             int end = s.back() - 'a';
             vvstr[beg][end].push_back(s);
         }
         DfsStart(vvstr);
         if(!ret.empty()){
             for (int i = ; i < ret.size(); i++) {
                 cout << ret[i] << " ";
             }
         }
         else {
             cout << "IMPOSSIBLE";
         }
         cout << endl;
     }

     return ;
 }