PAT 1087 All Roads Lead to Rome

  1 #include <cstdio>
  2 #include <climits>
  3 #include <iostream>
  4 #include <vector>
  5 #include <string>
  6 #include <queue>
  7 #include <unordered_map>
  8 #include <algorithm>
  9
 10 using namespace std;
 11
 12 typedef pair<int, int> P;
 13
 14 class City {
 15 public:
 16     int        id;
 17     string     name;
 18     int     happiness;
 19     int     distance;
 20     vector<int> adj;
 21     City(string _name, int _happiness = 0, int _id = 0, int _distance = INT_MAX) :
 22         name(_name), happiness(_happiness), id(_id), distance(_distance) {}
 23 };
 24
 25 class Solution {
 26 public:
 27     int happiness;
 28     vector<int> path;
 29     Solution(int happy = 0) : happiness(happy) {}
 30 };
 31
 32 bool solution_cmp_inv(const Solution& a, const Solution& b) {
 33     if (a.happiness > b.happiness) {
 34         return true;
 35     } else if (a.happiness < b.happiness) {
 36         return false;
 37     } else {
 38         return a.happiness / a.path.size() > b.happiness / b.path.size();
 39     }
 40 }
 41
 42 int G[200][200];
 43
 44 vector<Solution> res;
 45
 46 void build_solution(vector<City*> &cities, Solution &s, int pos, int dst, int hap) {
 47     // we know that start city index must be zero
 48     // so when pos == 0, we get a solution
 49     if (pos == 0) {
 50         res.push_back(s);
 51         res.back().happiness= hap;
 52         return;
 53     }
 54
 55     // we are now at the cities[pos] with left dst to the start city
 56     City* city = cities[pos];
 57     for (int i=0; i<city->adj.size(); i++) {
 58         City* adj = cities[city->adj[i]];
 59
 60         // we could not go to this adj city, not on the shortest path
 61         if (adj->distance + G[pos][adj->id] != dst) {
 62             continue;
 63         }
 64
 65         // here we can assume that the adj city is on the shortest path
 66         s.path.push_back(adj->id);
 67         build_solution(cities, s, adj->id, adj->distance, hap + adj->happiness);
 68         s.path.pop_back();
 69     }
 70 }
 71 int main() {
 72     vector<City*> cities;
 73
 74     unordered_map<string, int> city_lookup;
 75
 76     int        N, K;
 77     char    buf[10], buf2[10];
 78
 79     scanf("%d%d%s", &N, &K, buf);
 80
 81     City* start = new City(buf, 0, 0, 0);
 82     cities.push_back(start);
 83     city_lookup.insert(make_pair(start->name, 0));
 84
 85     int tmp = 0;
 86     for (int i=1; i<N; i++) {
 87         scanf("%s%d", buf, &tmp);
 88         cities.push_back(new City(buf, tmp, cities.size()));
 89         city_lookup.insert(make_pair(buf, cities.back()->id));
 90     }
 91
 92     tmp = 0;
 93     int ca = 0, cb = 0;
 94
 95     for (int i=0; i<K; i++) {
 96         scanf("%s%s%d", buf, buf2, &tmp);
 97
 98         ca = city_lookup.find(buf)->second;
 99         cb = city_lookup.find(buf2)->second;
100
101         cities[ca]->adj.push_back(cb);
102         cities[cb]->adj.push_back(ca);
103
104         G[ca][cb] = G[cb][ca] =tmp;
105     }
106
107     priority_queue<P, vector<P>, greater<P>> cand_cities;
108     cand_cities.push(make_pair(0, 0));
109
110     while(!cand_cities.empty()) {
111         P p = cand_cities.top();
112         cand_cities.pop();
113
114         City* sel_city = cities[p.second];
115
116         if (sel_city->distance < p.first) continue;
117
118         for (int i=0; i<sel_city->adj.size(); i++) {
119             City* adj_city = cities[sel_city->adj[i]];
120
121             int new_dst = sel_city->distance + G[sel_city->id][adj_city->id];
122
123             if (adj_city->distance > new_dst) {
124                 adj_city->distance = new_dst;
125                 cand_cities.push(make_pair(new_dst, adj_city->id));
126             }
127         }
128     }
129
130     City* rom = cities[city_lookup.find("ROM")->second];
131
132     Solution s;
133
134     build_solution(cities, s, rom->id, rom->distance, rom->happiness);
135
136     sort(res.begin(), res.end(), solution_cmp_inv);
137
138     Solution& best = res[0];
139
140     printf("%d %d %d %d\n", res.size(), rom->distance, best.happiness, best.happiness / best.path.size());
141
142     for (int i = best.path.size() - 1; i>=0; i--) {
143         printf("%s->", cities[best.path[i]]->name.c_str());
144     }
145     printf("ROM\n");
146     return 0;
147 }

稍微有点烦, 不过思路比较简单,还是跟课本靠的比较近最短路径,然后再dfs扫出所有可能路径, 进行一个排序就ok了,感谢STL,用单纯的C写的话...