PAT1053 

  1 #include<iostream>
  2 #include<vector>
  3 #include<stack>
  4 #include<algorithm>
  5 #include<iterator>
  6 using namespace std;
  7 
  8 #define MAX 9999
  9 
 10 void record(stack<int> st, vector<vector<int>> &paths)
 11 {
 12     vector<int> p;
 13     while(!st.empty())
 14     {
 15         p.push_back(st.top());
 16         st.pop();
 17     }
 18     reverse(p.begin(), p.end());
 19     paths.push_back(p);
 20 }
 21 
 22 void DFS_visit(vector<vector<int>> &tree, int S, int start, 
 23     vector<vector<int>> &paths, stack<int> &st, int sum, vector<int> &weight)
 24 {
 25     st.push(weight[start]);
 26     sum += weight[start];
 27     /*累计权重小于等于S进入判断下一个DFS，
 28     若大于则不用后续DFS*/
 29     if(sum <= S)
 30     {
 31         /*判断是否是叶子结点*/
 32         bool flag(false);
 33         for(int i=0; i<tree.size(); ++i)
 34             if(tree[start][i] != MAX)
 35             {
 36                 DFS_visit(tree, S, i, paths, st, sum, weight);
 37                 flag = true;
 38             }
 39         if(flag == false)
 40             if(sum == S)
 41                 record(st, paths);
 42     }
 43     sum -= st.top();
 44     st.pop();
 45 }
 46 
 47 void DFS_version(vector<vector<int>> &tree, int S, int start, 
 48     vector<vector<int>> &paths, vector<int> &weight)
 49 {
 50     stack<int> st;
 51     /*记录累加权重*/
 52     int sum(0);
 53     DFS_visit(tree, S, start, paths, st, sum, weight);
 54 }
 55 
 56 bool comp(vector<int> v1, vector<int> v2)
 57 {
 58     if(v1 == v2)
 59         return false;
 60     vector<int>::iterator iter_v1 = v1.begin();
 61     vector<int>::iterator iter_v2 = v2.begin();
 62     while(1)
 63         if((*iter_v1) > (*iter_v2))
 64             return true;
 65         else if((*iter_v1) < (*iter_v2))
 66             return false;
 67         else 
 68             {++iter_v1; ++iter_v2;}                                                                                                                                                                                                                                                                                                                                                   
 69 }
 70 
 71 int main()
 72 {
 73     int N, M, S;
 74     while(cin>>N>>M>>S)
 75     {
 76         vector<int> colum(N, MAX);
 77         /*存成有向图*/
 78         vector<vector<int>> tree(N, colum);
 79         vector<int> weight(N, 0);
 80         for(int i=0; i<N; ++i)
 81             cin>>weight[i];
 82         for(int i=0; i<M; ++i)
 83         {
 84             int id, k; cin>>id>>k;
 85             for(int j=0; j<k; ++j)
 86             {
 87                 int n; cin>>n;
 88                 tree[id][n] = 1;
 89             }
 90         }
 91         /*存储结果路径*/
 92         vector<vector<int>> paths;
 93         DFS_version(tree, S, 0, paths, weight);
 94         sort(paths.begin(), paths.end(), comp);
 95         for(int i=0; i<paths.size(); ++i)
 96         {
 97             for(int j=0; j<paths[i].size()-1; ++j)
 98                 cout<<paths[i][j]<<" ";
 99             cout<<paths[i][paths[i].size()-1]<<endl;
100         }
101     }
102     return 0;
103 }

 

题目链接：http://pat.zju.edu.cn/contests/pat-a-practise/1053

思路; 将树存成有向图，用DFS，配一个栈记录路径，最后对多条结果路径排序下，比较简单！

 

 

posted @ 2013-10-13 19:14  coding_monkey  阅读(323)  评论(0)  编辑  收藏  举报