《数据结构与算法分析：C语言描述》复习——第十章“算法设计技巧”——收费站重建问题

  1 // Turnpike reconstruction problem.
  2 //    Description:
  3 //        Given n * (n - 1) / 2 distances, find out if you can determine the 
  4 //            relative coordinates of n points.
  5 #include <algorithm>
  6 #include <cmath>
  7 #include <iostream>
  8 #include <map>
  9 #include <vector>
 10 using namespace std;
 11 
 12 int myabs(int x)
 13 {
 14     return x > 0 ? x : -x;
 15 }
 16 
 17 bool turnpikeReconstruction(int idx, vector<int> &x, map<int, int> &dist_set, 
 18     const int far)
 19 {
 20     if (idx == 0) {
 21         return true;
 22     }
 23     
 24     int cur_n;
 25     int cur_far;
 26     int i;
 27     map<int, int>::iterator it;
 28     
 29     cur_n = (int)x.size();
 30     cur_far = dist_set.rbegin()->first;
 31     for (i = 0; i < cur_n; ++i) {
 32         it = dist_set.find(myabs(cur_far - x[i]));
 33         if (it == dist_set.end() || it->second == 0) {
 34             break;
 35         }
 36         --it->second;
 37         if (it->second == 0) {
 38             dist_set.erase(it);
 39         }
 40     }
 41     if (i == cur_n) {
 42         x.push_back(cur_far);
 43         if (turnpikeReconstruction(idx - 1, x, dist_set, far)) {
 44             return true;
 45         }
 46         x.pop_back();
 47     }
 48     cur_n = i;
 49     for (i = 0; i < cur_n; ++i) {
 50         ++dist_set[myabs(cur_far - x[i])];
 51     }
 52     
 53     cur_n = (int)x.size();
 54     cur_far = dist_set.rbegin()->first;
 55     for (i = 0; i < cur_n; ++i) {
 56         it = dist_set.find(myabs(far - cur_far - x[i]));
 57         if (it == dist_set.end() || it->second == 0) {
 58             break;
 59         }
 60         --it->second;
 61         if (it->second == 0) {
 62             dist_set.erase(it);
 63         }
 64     }
 65     if (i == cur_n) {
 66         x.push_back(far - cur_far);
 67         if (turnpikeReconstruction(idx - 1, x, dist_set, far)) {
 68             return true;
 69         }
 70         x.pop_back();
 71     }
 72     cur_n = i;
 73     for (i = 0; i < cur_n; ++i) {
 74         ++dist_set[myabs(far - cur_far - x[i])];
 75     }
 76     
 77     return false;
 78 }
 79 
 80 int main()
 81 {
 82     int i;
 83     int n, n2;
 84     vector<int> x;
 85     int dist;
 86     map<int, int> dist_set;
 87     int far;
 88     
 89     while (cin >> n2 && n2 > 0) {
 90         for (i = 0; i < n2; ++i) {
 91             cin >> dist;
 92             ++dist_set[dist];
 93         }
 94         n = (int)sqrt(n2 * 2.0) + 1;
 95         far = dist_set.rbegin()->first;
 96         --dist_set[far];
 97         if (dist_set.rbegin()->second == 0) {
 98             dist_set.erase(far);
 99         }
100         
101         x.push_back(0);
102         x.push_back(far);
103         if (!turnpikeReconstruction(n - 2, x, dist_set, far)) {
104             cout << "No solution." << endl;
105         } else {
106             sort(x.begin(), x.end());
107             for (i = 0; i < n; ++i) {
108                 cout << x[i] << ‘ ‘;
109             }
110             cout << endl;
111         }
112         
113         x.clear();
114         dist_set.clear();
115     }
116     
117     return 0;
118 }