PAT_A1111#Online Map

Posted 2022-01-16 blue-lin

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:

  1 /*
  2 Data: 2019-06-16 17:40:10
  3 Problem: PAT_A1111#Online Map
  4 AC: 42:06
  5 
  6 题目大意：
  7 给出当前位置和目的地，给出最短路径和最快路径
  8 输入：
  9 第一行给出，结点数N，边数M
 10 接下来M行，v1,v2,单/双向，距离，时间
 11 最后一行，起点，终点
 12 输出：
 13 最短距离，输出路径，不唯一则选择最快的
 14 最短时间，输出路径，不唯一则选择经过结点最少的
 15 */
 16 
 17 #include<cstdio>
 18 #include<vector>
 19 #include<algorithm>
 20 using namespace std;
 21 const int M=1e3,INF=1e9;
 22 int grap[M][M],cost[M][M],vis[M],d[M];
 23 int n,v1,v2,optValue=INF,optCost=INF;
 24 vector<int> pre[M],temp,opt,pro[M],fav;
 25 
 26 void DijskraP(int s)
 27 
 28     fill(vis,vis+M,0);
 29     fill(d,d+M,INF);
 30     d[s]=0;
 31     for(int i=0; i<n; i++)
 32     
 33         int u=-1,Min=INF;
 34         for(int j=0; j<n; j++)
 35         
 36             if(vis[j]==0 && d[j]<Min)
 37             
 38                 Min=d[j];
 39                 u=j;
 40             
 41         
 42         if(u==-1)
 43             return;
 44         vis[u]=1;
 45         for(int v=0; v<n; v++)
 46         
 47             if(vis[v]==0 && grap[u][v]!=INF)
 48             
 49                 if(d[u]+grap[u][v] < d[v])
 50                 
 51                     d[v] = d[u]+grap[u][v];
 52                     pre[v].clear();
 53                     pre[v].push_back(u);
 54                 
 55                 else if(d[u]+grap[u][v] == d[v])
 56                     pre[v].push_back(u);
 57             
 58         
 59     
 60 
 61 
 62 void DFSP(int v)
 63 
 64     if(v == v1)
 65     
 66         temp.push_back(v);
 67         int value=0;
 68         for(int i=temp.size()-1; i>0; i--)
 69         
 70             int id=temp[i],idNext=temp[i-1];
 71             value += cost[id][idNext];
 72         
 73         if(value < optValue)
 74         
 75             optValue = value;
 76             opt = temp;
 77         
 78         temp.pop_back();
 79         return;
 80     
 81     temp.push_back(v);
 82     for(int i=0; i<pre[v].size(); i++)
 83         DFSP(pre[v][i]);
 84     temp.pop_back();
 85 
 86 
 87 void DijskraC(int s)
 88 
 89     fill(vis,vis+M,0);
 90     fill(d,d+M,INF);
 91     d[s]=0;
 92     for(int i=0; i<n; i++)
 93     
 94         int u=-1,Min=INF;
 95         for(int j=0; j<n; j++)
 96         
 97             if(vis[j]==0 && d[j]<Min)
 98             
 99                 Min=d[j];
100                 u=j;
101             
102         
103         if(u==-1)
104             return;
105         vis[u]=1;
106         for(int v=0; v<n; v++)
107         
108             if(vis[v]==0 && cost[u][v]!=INF)
109             
110                 if(d[u]+cost[u][v] < d[v])
111                 
112                     d[v] = d[u]+cost[u][v];
113                     pro[v].clear();
114                     pro[v].push_back(u);
115                 
116                 else if(d[u]+cost[u][v] == d[v])
117                     pro[v].push_back(u);
118             
119         
120     
121 
122 
123 void DFSC(int v)
124 
125     if(v == v1)
126     
127         temp.push_back(v);
128         if(temp.size() < optCost)
129         
130             optCost = temp.size();
131             fav = temp;
132         
133         temp.pop_back();
134         return;
135     
136     temp.push_back(v);
137     for(int i=0; i<pro[v].size(); i++)
138         DFSC(pro[v][i]);
139     temp.pop_back();
140 
141 
142 int main()
143 
144 #ifdef    ONLINE_JUDGE
145 #else
146     freopen("Test.txt", "r", stdin);
147 #endif
148 
149     fill(grap[0],grap[0]+M*M,INF);
150     fill(cost[0],cost[0]+M*M,INF);
151 
152     int m,one;
153     scanf("%d%d", &n,&m);
154     for(int i=0; i<m; i++)
155     
156         scanf("%d%d%d", &v1,&v2,&one);
157         scanf("%d%d", &grap[v1][v2],&cost[v1][v2]);
158         if(!one)
159         
160             grap[v2][v1]=grap[v1][v2];
161             cost[v2][v1]=cost[v1][v2];
162         
163     
164     scanf("%d%d",&v1,&v2);
165     DijskraP(v1);
166     DFSP(v2);
167     printf("Distance = %d", d[v2]);
168     DijskraC(v1);
169     DFSC(v2);
170     if(opt == fav)
171     
172         printf("; Time = %d: %d",d[v2],opt[opt.size()-1]);
173         for(int i=opt.size()-2; i>=0; i--)
174             printf(" -> %d", opt[i]);
175     
176     else
177     
178         printf(": %d", opt[opt.size()-1]);
179         for(int i=opt.size()-2; i>=0; i--)
180             printf(" -> %d", opt[i]);
181         printf("\n");
182         printf("Time = %d: %d", d[v2],fav[fav.size()-1]);
183         for(int i=fav.size()-2; i>=0; i--)
184             printf(" -> %d", fav[i]);
185     
186 
187 
188     return 0;
189