1111. Online Map (30)
Input our current position and a destination, an online map can recommend several paths. Now your job is to recommend two paths to your user: one is the shortest, and the other is the fastest. It is guaranteed that a path exists for any request.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers N (2 <= N <= 500), and M, being the total number of streets intersections on a map, and the number of streets, respectively. Then M lines follow, each describes a street in the format:
V1 V2 one-way length time
where V1 and V2 are the indices (from 0 to N-1) of the two ends of the street; one-way is 1 if the street is one-way from V1 to V2, or 0 if not; length is the length of the street; and time is the time taken to pass the street.
Finally a pair of source and destination is given.
Output Specification:
For each case, first print the shortest path from the source to the destination with distance D in the format:
Distance = D: source -> v1 -> ... -> destination
Then in the next line print the fastest path with total time T:
Time = T: source -> w1 -> ... -> destination
In case the shortest path is not unique, output the fastest one among the shortest paths, which is guaranteed to be unique. In case the fastest path is not unique, output the one that passes through the fewest intersections, which is guaranteed to be unique.
In case the shortest and the fastest paths are identical, print them in one line in the format:
Distance = D; Time = T: source -> u1 -> ... -> destination
Sample Input 1:
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
3 4 0 3 2
3 9 1 4 1
0 6 0 1 1
7 5 1 2 1
8 5 1 2 1
2 3 0 2 2
2 1 1 1 1
1 3 0 3 1
1 4 0 1 1
9 7 1 3 1
5 1 0 5 2
6 5 1 1 2
3 5
Sample Output 1:
Distance = 6: 3 -> 4 -> 8 -> 5
Time = 3: 3 -> 1 -> 5
Sample Input 2:
7 9
0 4 1 1 1
1 6 1 1 3
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 1 3
3 2 1 1 2
4 5 0 2 2
6 5 1 1 2
3 5
Sample Output 2:
Distance = 3; Time = 4: 3 -> 2 -> 5
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = , inf = 0x7fffffff;
struct Arc{
int v;
int tim, len;
};
vector<Arc> arc[maxn];
int N, M, dis[maxn], tim[maxn], pre[maxn], pathLen[maxn], S, T;
vector<int> disPath, timPath, tempPath; void scan(){
scanf("%d%d", &N, &M);
for(int i = ; i < M; ++ i){
int v1, v2, oneWay;
Arc t;
scanf("%d%d%d%d%d", &v1, &v2, &oneWay, &t.len, &t.tim);
t.v = v2; arc[v1].push_back(t);
if(!oneWay) t.v = v1, arc[v2].push_back(t);
}
scanf("%d%d", &S, &T);
} void dijkstra_dis(int s){
bool vis[maxn];
fill(vis, vis+maxn, false);
fill(dis, dis+maxn, inf);
fill(tim, tim+maxn, inf);
dis[s] = tim[s] = ;
for(int i = ; i < N; ++ i){
int u = -, minDis = inf;
for(int j = ; j < N; ++ j){
if(!vis[j] && dis[j] < minDis){
minDis = dis[j];
u = j;
}
}
if(u == -) return;
vis[u] = true;
for(int k = ; k < arc[u].size(); ++ k){
int v = arc[u][k].v;
if(!vis[v]){
if(dis[u] + arc[u][k].len < dis[v]){
dis[v] = dis[u] + arc[u][k].len;
tim[v] = tim[u] + arc[u][k].tim;
pre[v] = u;
}else if(dis[u] + arc[u][k].len == dis[v] && tim[u] + arc[u][k].tim < tim[v]){
tim[v] = tim[u] + arc[u][k].tim;
pre[v] = u;
}
}
}
}
} void dijkstra_tim(int s){
bool vis[maxn];
fill(vis, vis+maxn, false);
fill(tim, tim+maxn, inf);
fill(pathLen, pathLen+maxn, inf);
tim[s] = , pathLen[s] = ;
for(int i = ; i < N; i ++){
int u = -, minTim = inf;
for(int j = ; j < N; j ++){
if(!vis[j] && tim[j] < minTim){
minTim = tim[j];
u = j;
}
}
if(u == -) return;
vis[u] = true;
for(int k = ; k < arc[u].size(); k ++){
int v = arc[u][k].v;
if(!vis[v]){
if(tim[u] + arc[u][k].tim < tim[v]){
tim[v] = tim[u] + arc[u][k].tim;
pre[v] = u;
pathLen[v] = pathLen[u] + ;
}else if(tim[u] + arc[u][k].tim == tim[v] && pathLen[u]+ < pathLen[v]){
pre[v] = u;
pathLen[v] = pathLen[u] + ;
}
}
}
}
} void dfs(int t){
tempPath.push_back(t);
if(pre[t] != -){
dfs(pre[t]);
}
} void printPath(vector<int> &path){
for(int i = path.size()-; i >= ; -- i){
printf("%d", path[i]);
if(i != ) printf(" -> ");
}
printf("\n");
} int main()
{
fill(pre, pre+maxn, -);
scan();
dijkstra_dis(S);
dfs(T);
disPath = tempPath;
fill(pre, pre+maxn, -);
dijkstra_tim(S);
tempPath.clear();//clear data
dfs(T);
timPath = tempPath;
if(timPath == disPath){
printf("Distance = %d; Time = %d: ", dis[T], tim[T]);
printPath(disPath);
}else{
printf("Distance = %d: ", dis[T]); printPath(disPath);
printf("Time = %d: ", tim[T]); printPath(timPath);
}
return ;
}
1111. Online Map (30)的更多相关文章
- PAT (Advanced Level) 1111. Online Map (30)
预处理出最短路再进行暴力dfs求答案会比较好.直接dfs效率太低. #include<cstdio> #include<cstring> #include<cmath&g ...
- 1111 Online Map (30)(30 分)
Input our current position and a destination, an online map can recommend several paths. Now your jo ...
- PAT Advanced 1111 Online Map (30) [Dijkstra算法 + DFS]
题目 Input our current position and a destination, an online map can recommend several paths. Now your ...
- PAT甲题题解-1111. Online Map (30)-PAT甲级真题(模板题,两次Dijkstra,同时记下最短路径)
题意:给了图,以及s和t,让你求s到t花费的最短路程.最短时间,以及输出对应的路径. 对于最短路程,如果路程一样,输出时间最少的. 对于最短时间,如果时间一样,输出节点数最少的. 如果最短路程 ...
- 【PAT甲级】1111 Online Map (30分)(dijkstra+路径记录)
题意: 输入两个正整数N和M(N<=500,M<=N^2),分别代表点数和边数.接着输入M行每行包括一条边的两个结点(0~N-1),这条路的长度和通过这条路所需要的时间.接着输入两个整数表 ...
- 1111 Online Map (30 分)
1111. Online Map (30)Input our current position and a destination, an online map can recommend sever ...
- PAT甲级——1111 Online Map (单源最短路经的Dijkstra算法、priority_queue的使用)
本文章同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90041078 1111 Online Map (30 分) ...
- 1111 Online Map (30 分)
1111 Online Map (30 分) Input our current position and a destination, an online map can recommend sev ...
- PAT 1111 Online Map[Dijkstra][dfs]
1111 Online Map(30 分) Input our current position and a destination, an online map can recommend seve ...
随机推荐
- systemctl启动tomcat后,jps看不到进程
centos7 写了tomcat的启动脚本,脚本从网上copy的. [Unit]Description=tomcat-1After=syslog.target network.target remot ...
- Unity 自定义导入时切割Sprite
using System.Collections; using System.Collections.Generic; using UnityEngine; using UnityEditor; us ...
- Windows邮件客户端
换回WIndows 因为要保存邮件 所以需要邮件客户端 Foxmail 腾讯自家 同样需要独立密码 之前的foxmail是全拼 新注册了一个@qq 发现新注册的@qq绑定到的是新的QQ号 算了 去用1 ...
- CAS 无锁队列
队列是常用的数据结构,采用的FIFO(first in firstout)原则,新元素(等待进入队列的元素)总是被插入到尾部,而读取的时候总是从头部开始读取.在计算中队列一般用来做排队(如线程池的等待 ...
- 消息中间件 rabbitMQ
参考内容:https://www.cnblogs.com/dwlsxj/p/RabbitMQ.html 一.什么是RabbitMQ RabbitMQ是一个由erlang 开发的AMQP(Advance ...
- 使用电脑ODBC测试数据库连接方法
使用电脑ODBC测试数据库连接方法 一.打开电脑的控制面板——管理工具——数据源(ODBC),在用户dsn页面中点击添加按钮,选择IBM DB2 ODBC DRIVER,点击完成. 二.在弹出的配置页 ...
- sqlserver 收缩数据库/文件
/******************************/ 1.右键-属性-选项-简单模式 2.右键-任务-收缩-文件 3.右键-任务-收缩-数据库 /********************* ...
- ExpandableListView解析JSON数据
效果图: 说明:刚开始使用这个控件我花费了3天的时间,但是一直都没有达到预期的效果,要么就是直接全部不显示,要么就是数据累加 ...
- innodb 体系结构(后台进程)
一.后台进程(innodb 1.0.x以前的) 1.master thread master thread具有最高的线程优先级别,其内部由多个循环(loop)组成:主循环(loop).后台循环(bac ...
- WPF 初识
1.WPF 与Winform比较 1.1.WPF所有的操作都不依赖于GDI和GDI+,而是间接依赖于强大的Direct3D,这就意味着通过WPF可以做出以前WinFrom无法想象的视觉效果,包括3D效 ...