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

知识点： Dijkstra算法； DFS算法

思路：

第一是求最短路径，有相同的则输出时间最短；利用Dijkstra求解，更新最短路时，如果最短路相同，则比较时间

1 if(!visited[i]){ 2     if(minl[i]>minD+G_l[minV][i]){ 3         minl[i]=minD+G_l[minV][i]; 4         mint[i]=mint[minV]+G_t[minV][i]; 5         pre_shorest[i]=minV; 6     }else if(minl[i]==minD+G_l[minV][i]&& 7             mint[i]>mint[minV]+G_t[minV][i]){ 8         mint[i]=mint[minV]+G_t[minV][i]; 9         pre_shorest[i]=minV;        10     }11 }

第二是求最快路，如果有相同的，输出节点最少的：用Dijkstra算法，设立容器pre来储存每个节点的优选前去节点；然后用DFS来遍历每条路径，选出最少节点的

1 for(int i=0;i
     
      minD+G_t[minV][i]){ 4             mint[i]=mint[minV]+G_t[minV][i]; 5             pre_faster[i].clear(); 6             pre_faster[i].push_back(minV); 7         }else if(mint[i]==minD+G_t[minV][i]){ 8             pre_faster[i].push_back(minV);         9         }10     }11 }
     

 

最后，vector可以比较，相同的情况特殊处理

 

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 
     
      #include 
      
       #include 
       
        using namespace std;const int maxn = 550;const int inf = 999999;int n,m;int G_l[maxn][maxn];int G_t[maxn][maxn];int minl[maxn];int mint[maxn];int visited[maxn];int pre_shorest[maxn];vector
        
          pre_faster[maxn];vector
         
           s_path;vector
          
            f_path;vector
           
             f_tmpp;int minsize;int dijkstra_shorest(int start,int end){ fill(pre_shorest, pre_shorest+maxn, -1); fill(minl,minl+maxn,inf); fill(mint,mint+maxn,inf); fill(visited,visited+maxn,0); minl[start] = 0; for(int i=0;i
            
             minD+G_l[minV][i]){ minl[i]=minD+G_l[minV][i]; mint[i]=mint[minV]+G_t[minV][i]; pre_shorest[i]=minV; }else if(minl[i]==minD+G_l[minV][i]&& mint[i]>mint[minV]+G_t[minV][i]){ mint[i]=mint[minV]+G_t[minV][i]; pre_shorest[i]=minV; } } } } int ptr = end; while(ptr != -1){ //printf(" %d\n",ptr); s_path.push_back(ptr); ptr=pre_shorest[ptr]; } return minl[end];}void DFS(int v,int start){ f_tmpp.push_back(v); if(v==start){ if(f_tmpp.size()
             
              minD+G_t[minV][i]){ mint[i]=mint[minV]+G_t[minV][i]; pre_faster[i].clear(); pre_faster[i].push_back(minV); }else if(mint[i]==minD+G_t[minV][i]){ pre_faster[i].push_back(minV); } } } } minsize = inf; DFS(end,start); for(int i=0;i
              
               =0;i--){ printf("%d",s_path[i]); if(i!=0) printf(" -> "); } }else{ printf("Distance = %d: ",D); for(int i=s_path.size()-1;i>=0;i--){ printf("%d",s_path[i]); if(i!=0) printf(" -> "); } printf("\nTime = %d: ",T); for(int i=f_path.size()-1;i>=0;i--){ printf("%d",f_path[i]); if(i!=0) printf(" -> "); } }}