PAT_A1111#Online Map

Source:

PAT A1111 Online Map (30 分)

Description:

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), 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

Keys:

• 最短路径

Attention:

• 其实就是求两次最短路径，代码code一遍再copy一遍，考试的时候怎么快怎么来，优化代码反而容易出错

Code:

 /*
 Data: 2019-08-18 21:30:18
 Problem: PAT_A1111#Online Map
 AC: 42:06

 题目大意：
 给出当前位置和目的地，给出最短路径和最快路径
 输入：
 第一行给出，结点数N，边数M
 接下来M行，v1,v2,单/双向，距离，时间
 最后一行，起点，终点
 输出：
 最短距离，输出路径，不唯一则选择最快的
 最短时间，输出路径，不唯一则选择经过结点最少的
 */
 #include<cstdio>
 #include<vector>
 #include<algorithm>
 using namespace std;
 const int M=1e3,INF=1e9;
 int grap[M][M],cost[M][M],vis[M],d[M],c[M];
 int n,m,v1,v2,one,optTime=INF,optLeth=INF;
 vector<int> preL[M],preT[M],optL,pathL,optT,pathT;

 void DijskraL(int s)
 {
     fill(vis,vis+M,);
     fill(d,d+M,INF);
     d[s]=;
     for(int i=; i<n; i++)
     {
         int u=-,Min=INF;
         for(int j=; j<n; j++)
         {
             if(vis[j]== && d[j]<Min)
             {
                 Min = d[j];
                 u = j;
             }
         }
         if(u==-)   return;
         vis[u]=;
         for(int v=; v<n; v++)
         {
             if(vis[v]== && grap[u][v]!=INF)
             {
                 if(d[u]+grap[u][v] < d[v])
                 {
                     preL[v].clear();
                     preL[v].push_back(u);
                     d[v] = d[u]+grap[u][v];
                 }
                 else if(d[u]+grap[u][v]==d[v])
                     preL[v].push_back(u);
             }
         }
     }
 }

 void DFSL(int s)
 {
     if(s == v1)
     {
         pathL.push_back(v1);
         int time=,len=pathL.size();
         for(int i=; i<len; i++)
             time += cost[pathL[i]][pathL[i-]];
         if(time < optTime)
         {
             optTime = time;
             optL = pathL;
         }
         pathL.pop_back();
         return;
     }
     pathL.push_back(s);
     for(int i=; i<preL[s].size(); i++)
         DFSL(preL[s][i]);
     pathL.pop_back();
 }

 void DijskraT(int s)
 {
     fill(vis,vis+M,);
     fill(c,c+M,INF);
     c[s]=;
     for(int i=; i<n; i++)
     {
         int u=-,Min=INF;
         for(int j=; j<n; j++)
         {
             if(vis[j]== && c[j]<Min)
             {
                 Min = c[j];
                 u = j;
             }
         }
         if(u==-)   return;
         vis[u]=;
         for(int v=; v<n; v++)
         {
             if(vis[v]== && cost[u][v]!=INF)
             {
                 if(c[u]+cost[u][v] < c[v])
                 {
                     preT[v].clear();
                     preT[v].push_back(u);
                     c[v] = c[u] + cost[u][v];
                 }
                 else if(c[u]+cost[u][v]==c[v])
                     preT[v].push_back(u);
             }
         }
     }
 }

 void DFST(int s)
 {
     if(s == v1)
     {
         pathT.push_back(v1);
         if(pathT.size() < optLeth)
         {
             optLeth = pathT.size();
             optT = pathT;
         }
         pathT.pop_back();
         return;
     }
     pathT.push_back(s);
     for(int i=; i<preT[s].size(); i++)
         DFST(preT[s][i]);
     pathT.pop_back();
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     fill(grap[],grap[]+M*M,INF);
     fill(cost[],cost[]+M*M,INF);

     scanf("%d%d", &n,&m);
     for(int i=; i<m; i++)
     {
         scanf("%d%d%d", &v1,&v2,&one);
         scanf("%d%d", &grap[v1][v2],&cost[v1][v2]);
         if(!one)
         {
             grap[v2][v1]=grap[v1][v2];
             cost[v2][v1]=cost[v1][v2];
         }
     }
     scanf("%d%d", &v1,&v2);
     DijskraL(v1);
     DFSL(v2);
     DijskraT(v1);
     DFST(v2);
     if(optT == optL)
     {
         printf("Distance = %d; Time = %d: %d", d[v2],c[v2],optT[optT.size()-]);
         for(int i=optT.size()-; i>=; i--)
             printf(" -> %d", optT[i]);
     }
     else
     {
         printf("Distance = %d: %d", d[v2], optL[optL.size()-]);
         for(int i=optL.size()-; i>=; i--)
             printf(" -> %d", optL[i]);
         printf("\nTime = %d: %d", c[v2], optT[optT.size()-]);
         for(int i=optT.size()-; i>=; i--)
             printf(" -> %d", optT[i]);
     }

     return ;
 }