PAT 1087
又是最短路，陈越姥姥好像很喜欢最短路... 这道题之后还得再写一遍，最短路一定要熟悉
碰到一个坑，就是在加最短路的时候，不是++，而是要把前面的最短路加进来...
1 #include <vector> 2 #include <string> 3 #include <iostream> 4 #include <map> 5 #include <fstream> 6 7 //#define OJ 8 9 #ifdef OJ 10 #define fin cin 11 #endif 12 13 using namespace std; 14 15 class city{ 16 public: 17 int happiness; 18 string name; 19 20 vector<int> adj; 21 vector<int> cost; 22 }; 23 24 vector<city> cities; 25 map<string, int> city_idx; 26 map<int, string> city_name; 27 28 void dij(){ 29 int total = cities.size(); 30 int MAX = 0x7fffffff; 31 32 vector<bool> visited(total, false); 33 vector<int> dij_dist(total, MAX); 34 vector<int> dij_happiness(total, 0); 35 vector<int> prev(total, -1); 36 vector<int> dij_cnt(total, MAX); 37 vector<int> dij_least_cnt(total, 0); 38 39 int target_idx = city_idx["ROM"]; 40 41 dij_dist[0] = 0; 42 dij_cnt[0] = 0; 43 dij_least_cnt[0] = 1; 44 45 while (true){ 46 int least_dist = MAX; 47 int least_idx = -1; 48 49 for (int i = 0; i < total; i++){ 50 if (!visited[i] && least_dist > dij_dist[i]){ 51 least_idx = i; 52 least_dist = dij_dist[i]; 53 } 54 } 55 56 if (least_idx == -1) 57 break; 58 if (least_idx == target_idx) // ? 59 break; 60 61 visited[least_idx] = true; 62 vector<int> &adj = cities[least_idx].adj; 63 vector<int> &cost = cities[least_idx].cost; 64 for (int i = 0; i < adj.size(); i++){ 65 int cur_idx = adj[i]; 66 if (visited[cur_idx]) 67 continue; 68 if (least_dist + cost[i] < dij_dist[cur_idx]){ 69 dij_dist[cur_idx] = least_dist + cost[i]; 70 prev[cur_idx] = least_idx; 71 dij_cnt[cur_idx] = dij_cnt[least_idx] + 1; 72 dij_happiness[cur_idx] = dij_happiness[least_idx] + cities[cur_idx].happiness; 73 dij_least_cnt[cur_idx] = dij_least_cnt[least_idx]; 74 } 75 else if (least_dist + cost[i] == dij_dist[cur_idx]){ 76 dij_least_cnt[cur_idx] += dij_least_cnt[least_idx]; 77 78 if (dij_happiness[least_idx] + cities[cur_idx].happiness > dij_happiness[cur_idx]){ 79 prev[cur_idx] = least_idx; 80 dij_cnt[cur_idx] = dij_cnt[least_idx] + 1; 81 dij_happiness[cur_idx] = dij_happiness[least_idx] + cities[cur_idx].happiness; 82 } 83 else if (dij_happiness[least_idx] + cities[cur_idx].happiness == dij_happiness[cur_idx]){ 84 if (dij_cnt[least_idx] + 1 < dij_cnt[cur_idx]){ 85 prev[cur_idx] = least_idx; 86 dij_cnt[cur_idx] = dij_cnt[least_idx] + 1; 87 } 88 } 89 } 90 } 91 } 92 93 int idx = target_idx; 94 cout << dij_least_cnt[target_idx] << " " << dij_dist[target_idx] << " " << dij_happiness[target_idx] << " " << (int)(dij_happiness[target_idx] / dij_cnt[target_idx]) << endl; 95 96 vector<int> res; 97 98 while (idx != 0){ 99 res.push_back(idx); 100 idx = prev[idx]; 101 } 102 cout << city_name[0]; 103 for (int i = res.size() - 1; i >= 0; i--) 104 cout << "->" << city_name[res[i]]; 105 cout << endl; 106 } 107 108 int main(){ 109 #ifndef OJ 110 ifstream fin; 111 fin.open("in.data"); 112 #endif 113 114 115 int N, K; 116 fin >> N >> K; 117 118 string name; 119 fin >> name; 120 city_idx[name] = 0; 121 city_name[0] = name; 122 123 city c; 124 c.name = name; 125 c.happiness = 0; 126 127 cities.push_back(c); 128 129 for (int i = 1; i < N; i++){ 130 fin >> c.name; 131 fin >> c.happiness; 132 133 cities.push_back(c); 134 135 city_name[i] = c.name; 136 city_idx[c.name] = i; 137 } 138 139 for (int i = 0; i < K; i++){ 140 int c1, c2, cost; 141 142 fin >> name; 143 c1 = city_idx[name]; 144 145 fin >> name; 146 c2 = city_idx[name]; 147 148 fin >> cost; 149 150 cities[c1].adj.push_back(c2); 151 cities[c1].cost.push_back(cost); 152 cities[c2].adj.push_back(c1); 153 cities[c2].cost.push_back(cost); 154 } 155 156 dij(); 157 158 return 0; 159 }
