L3-007. 天梯地图
时间限制
300 ms
内存限制
65536 kB
代码长度限制
8000 B
判题程序
Standard
作者
陈越
本题要求你实现一个天梯赛专属在线地图,队员输入自己学校所在地和赛场地点后,该地图应该推荐两条路线:一条是最快到达路线;一条是最短距离的路线。题目保证对任意的查询请求,地图上都至少存在一条可达路线。
输入格式:
输入在第一行给出两个正整数N(2 <= N <=500)和M,分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行,每行按如下格式给出一条道路的信息:
V1 V2 one-way length time
其中V1和V2是道路的两个端点的编号(从0到N-1);如果该道路是从V1到V2的单行线,则one-way为1,否则为0;length是道路的长度;time是通过该路所需要的时间。最后给出一对起点和终点的编号。
输出格式:
首先按下列格式输出最快到达的时间T和用节点编号表示的路线:
Time = T: 起点 => 节点1 => ... => 终点
然后在下一行按下列格式输出最短距离D和用节点编号表示的路线:
Distance = D: 起点 => 节点1 => ... => 终点
如果最快到达路线不唯一,则输出几条最快路线中最短的那条,题目保证这条路线是唯一的。而如果最短距离的路线不唯一,则输出途径节点数最少的那条,题目保证这条路线是唯一的。
如果这两条路线是完全一样的,则按下列格式输出:
Time = T; Distance = D: 起点 => 节点1 => ... => 终点
输入样例1:10 15 0 1 0 1 1 8 0 0 1 1 4 8 1 1 1 5 4 0 2 3 5 9 1 1 4 0 6 0 1 1 7 3 1 1 2 8 3 1 1 2 2 5 0 2 2 2 1 1 1 1 1 5 0 1 3 1 4 0 1 1 9 7 1 1 3 3 1 0 2 5 6 3 1 2 1 5 3输出样例1:
Time = 6: 5 => 4 => 8 => 3 Distance = 3: 5 => 1 => 3输入样例2:
7 9 0 4 1 1 1 1 6 1 3 1 2 6 1 1 1 2 5 1 2 2 3 0 0 1 1 3 1 1 3 1 3 2 1 2 1 4 5 0 2 2 6 5 1 2 1 3 5输出样例2:
Time = 3; Distance = 4: 3 => 2 => 5
// 醉了,最后一个点死活过不了,以后想起来在说吧= = #include <iostream> #include <vector> #define INF 1 << 30 using namespace std; struct node { int end; int length; int time; node(int end, int length, int time) { this->end = end; this->length = length; this->time = time; } }; // 最短路线&&结点最少 vector<int> dijkstra_len(int start, int end, int N, vector< vector<node> >& map) { vector<int> dis_len(N, INF); // distance vector<bool> vis(N, false); // visit vector<int> path(N + 10, -1); // path 记录前导点 vector<int> point(N + 1, INF); dis_len[start] = 0; point[start] = 1; while (true) { int cur = -1; int min = INF; for (int i = 0; i < N; i++) { if (vis[i] != true && dis_len[i] < min) { cur = i; min = dis_len[i]; } } if (cur == -1) break; vis[cur] = true; for (int i = 0; i < map[cur].size(); i++) { if (vis[map[cur][i].end] == true) continue; // 对距离进行松弛 if (dis_len[map[cur][i].end] > dis_len[cur] + map[cur][i].length) { dis_len[map[cur][i].end] = dis_len[cur] + map[cur][i].length; point[map[cur][i].end] = point[cur] + 1; path[map[cur][i].end] = cur; } // 如果距离相同,则对结点数进行松弛 else if ((dis_len[map[cur][i].end] == dis_len[cur] + map[cur][i].length) && (point[map[cur][i].end] > point[cur] + 1)) { point[map[cur][i].end] = point[cur] + 1; path[map[cur][i].end] = cur; } } } // 返回路径,通过前导推出 vector<int> return_path; int cur = end; while (cur != -1) { return_path.push_back(cur); cur = path[cur]; } // 在路径后面添加一个最短距离,方便后续统计 return_path.push_back(dis_len[end]); return return_path; } // 时间最短&&路径最短 vector<int> dijkstra_time(int start, int end, int N, vector< vector<node> >& map) { vector<int> dis_time(N, INF); vector<int> dis_len(N, INF); vector<bool> vis(N, false); vector<int> path(N + 10, -1); dis_time[start] = 0; dis_len[start] = 0; while (true) { int cur = -1; int min = INF; for (int i = 0; i < N; i++) { if (vis[i] != true && dis_time[i] < min) { cur = i; min = dis_time[i]; } } if (cur == -1) break; vis[cur] = true; for (int i = 0; i < map[cur].size(); i++) { if (vis[map[cur][i].end] == true) continue; // 对时间进行松弛 if (dis_time[map[cur][i].end] > dis_time[cur] + map[cur][i].time) { dis_time[map[cur][i].end] = dis_time[cur] + map[cur][i].time; dis_len[map[cur][i].end] = dis_len[cur] + map[cur][i].length; path[map[cur][i].end] = cur; } // 如果时间相同,则对距离进行松弛 else if ((dis_time[map[cur][i].end] == dis_time[cur] + map[cur][i].time) && (dis_len[map[cur][i].end] > dis_len[cur] + map[cur][i].length)){ dis_len[map[cur][i].end] = dis_len[cur] + map[cur][i].length; path[map[cur][i].end] = cur; } } } vector<int> return_path; int cur = end; while (cur != -1) { return_path.push_back(cur); cur = path[cur]; } // 在路径后面添加一个最短距离,方便 return_path.push_back(dis_time[end]); // 返回最短路径 return return_path; } int main() { //freopen("1.txt", "r", stdin); int N, M; int start, end; cin >> N >> M; vector< vector<node> > map(N); vector< int > path_len; vector< int > path_time; for (int i = 0; i < M; i++) { int start, end, one_way, length, time; cin >> start >> end >> one_way >> length >> time; map[start].push_back(node(end, length, time)); if (one_way == 0) map[end].push_back(node(start, length, time)); } cin >> start >> end; path_len = dijkstra_len(start, end, N, map); path_time = dijkstra_time(start, end, N, map); bool flag = false; if (path_len.size() == path_time.size()) { for (int i = path_len.size() - 2; i >= 0; i--) if (path_len[i] != path_time[i]) break; flag = true; } else flag = false; if (false == flag) { printf("Time = %d: %d", path_time[path_time.size() - 1], path_time[path_time.size() - 2]); for (int i = path_time.size() - 3; i >= 0; i--) printf(" => %d", path_time[i]); printf("\n"); printf("Distance = %d: %d", path_len[path_len.size() - 1], path_len[path_len.size() - 2]); for (int i = path_len.size() - 3; i >= 0; i--) printf(" => %d", path_len[i]); } else { printf("Time = %d; Distance = %d: %d", path_time[path_time.size() - 1], path_len[path_len.size() - 1], path_len[path_len.size() - 2]); for (int i = path_len.size() - 3; i >= 0; i--) printf(" => %d", path_len[i]); } return 0; }