PAT1030:Travel Plan

1030. Travel Plan (30)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

A traveler's map gives the distances between cities along the highways, together with the cost of each highway. Now you are supposed to write a program to help a traveler to decide the shortest path between his/her starting city and the destination. If such a shortest path is not unique, you are supposed to output the one with the minimum cost, which is guaranteed to be unique.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers N, M, S, and D, where N (<=500) is the number of cities (and hence the cities are numbered from 0 to N-1); M is the number of highways; S and D are the starting and the destination cities, respectively. Then M lines follow, each provides the information of a highway, in the format:

City1 City2 Distance Cost

where the numbers are all integers no more than 500, and are separated by a space.

Output Specification:

For each test case, print in one line the cities along the shortest path from the starting point to the destination, followed by the total distance and the total cost of the path. The numbers must be separated by a space and there must be no extra space at the end of output.

Sample Input

Sample Output

0 2 3 3 40

思路

迪杰斯特拉单源最短路径算法和dfs的结合应用，比较绕，开始写的时候感觉比较难。概括来说主要为以下两步：

1.首先通过Dijkstra算法将起始点到所有点的最短距离计算出来，并用一个pre数组保存下所有路径节点的前驱结点（可能有多个前驱点，因为最短路径可能不止一条）。
2.从终点开始用dfs回溯到起点，因为最短路径不止一条，所以用一个tmp记录下回溯的每一条路径，然后通过比较路径的消耗总权值确定消耗最小的那条最短路径。

代码

#include<iostream>
#include<vector>
#include<math.h>
using namespace std;
const int Infmax = pow(2,30);
int N,M,S,D;
vector<vector<int>> graph(502,vector<int>(502,Infmax)); //The graph
vector<int> Distance(502,Infmax); //The shortest distance from start to every node(Dynamic Update)
vector<vector<int>> pre(502); //a node's previous nodes with a shortest path.
vector<bool> visit(502,false); // check if the node is visited
vector<vector<int>> cost(502,vector<int>(502,0)); //every road's cost
vector<int> path,tmp;//the final path and temp path
int mincost = Infmax;

void dfs(int v) //track from the destination to start and find the min cost.
{
   if(v == S)
   {
      tmp.push_back(v);
      int tmpcost = 0;
      for(int i = tmp.size() - 1;i > 0;i--)
      {
          tmpcost += cost[ tmp[i] ][ tmp[i - 1] ];
      }
      if(tmpcost < mincost)
      {
          mincost = tmpcost;
          path = tmp;
      }
      tmp.pop_back();
      return;
   }
   tmp.push_back(v);
   for(int i = 0;i < pre[v].size();i++) // dfs all of the previous nodes of this node
     dfs(pre[v][i]);
   tmp.pop_back();
}

int main()
{
   cin >> N >> M >> S >> D;
   for(int i = 0;i < M;i++) //Input
   {
       int s,d;
       cin >> s >> d;
       cin >> graph[s][d];
       graph[d][s] = graph[s][d];
       cin >> cost[s][d];
       cost[d][s] = cost[s][d];
   }

   //Confirm the start node
   pre[S].push_back(S);
   Distance[S] = 0;

   //Dijkstra's main algorithm
   for(int i = 0;i < N;i++)
   {
       int u = -1,nextmin = Infmax;
       for(int j = 0;j < N;j++)    //Find the current shortest path from current node to next node
       {
          if(!visit[j] && Distance[j] < nextmin)
          {
              u = j;
              nextmin = Distance[j];
          }
       }
       if(u == -1) break; //All of the Nodes have been visited
       visit[u] = true;
       for(int v = 0;v < N;v++) //search and update the shortest path of every node
       {
           if(!visit[v] && graph[u][v] != Infmax)
           {
               if(Distance[u] + graph[u][v] < Distance[v])
               {
                   Distance[v] = Distance[u] + graph[u][v];
                   pre[v].clear();
                   pre[v].push_back(u);
               }
               else if(Distance[u] + graph[u][v] == Distance[v])
               {
                   pre[v].push_back(u);
               }
           }
       }
   }

   dfs(D); //track back form the destination(D) to start(S) and find the min cost.
   //output
   for(int i = path.size()-1;i >=0;i--) //print the path
     cout << path[i] << " ";
   cout << Distance[D] << " " << mincost << endl; //print the shortest path's length and min cost
}