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:

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: 2019-08-18 21:30:18
  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 #include<cstdio>
 17 #include<vector>
 18 #include<algorithm>
 19 using namespace std;
 20 const int M=1e3,INF=1e9;
 21 int grap[M][M],cost[M][M],vis[M],d[M],c[M];
 22 int n,m,v1,v2,one,optTime=INF,optLeth=INF;
 23 vector<int> preL[M],preT[M],optL,pathL,optT,pathT;
 24 
 25 void DijskraL(int s)
 26 {
 27     fill(vis,vis+M,0);
 28     fill(d,d+M,INF);
 29     d[s]=0;
 30     for(int i=0; i<n; i++)
 31     {
 32         int u=-1,Min=INF;
 33         for(int j=0; j<n; j++)
 34         {
 35             if(vis[j]==0 && d[j]<Min)
 36             {
 37                 Min = d[j];
 38                 u = j;
 39             }
 40         }
 41         if(u==-1)   return;
 42         vis[u]=1;
 43         for(int v=0; v<n; v++)
 44         {
 45             if(vis[v]==0 && grap[u][v]!=INF)
 46             {
 47                 if(d[u]+grap[u][v] < d[v])
 48                 {
 49                     preL[v].clear();
 50                     preL[v].push_back(u);
 51                     d[v] = d[u]+grap[u][v];
 52                 }
 53                 else if(d[u]+grap[u][v]==d[v])
 54                     preL[v].push_back(u);
 55             }
 56         }
 57     }
 58 }
 59 
 60 void DFSL(int s)
 61 {
 62     if(s == v1)
 63     {
 64         pathL.push_back(v1);
 65         int time=0,len=pathL.size();
 66         for(int i=1; i<len; i++)
 67             time += cost[pathL[i]][pathL[i-1]];
 68         if(time < optTime)
 69         {
 70             optTime = time;
 71             optL = pathL;
 72         }
 73         pathL.pop_back();
 74         return;
 75     }
 76     pathL.push_back(s);
 77     for(int i=0; i<preL[s].size(); i++)
 78         DFSL(preL[s][i]);
 79     pathL.pop_back();
 80 }
 81 
 82 void DijskraT(int s)
 83 {
 84     fill(vis,vis+M,0);
 85     fill(c,c+M,INF);
 86     c[s]=0;
 87     for(int i=0; i<n; i++)
 88     {
 89         int u=-1,Min=INF;
 90         for(int j=0; j<n; j++)
 91         {
 92             if(vis[j]==0 && c[j]<Min)
 93             {
 94                 Min = c[j];
 95                 u = j;
 96             }
 97         }
 98         if(u==-1)   return;
 99         vis[u]=1;
100         for(int v=0; v<n; v++)
101         {
102             if(vis[v]==0 && cost[u][v]!=INF)
103             {
104                 if(c[u]+cost[u][v] < c[v])
105                 {
106                     preT[v].clear();
107                     preT[v].push_back(u);
108                     c[v] = c[u] + cost[u][v];
109                 }
110                 else if(c[u]+cost[u][v]==c[v])
111                     preT[v].push_back(u);
112             }
113         }
114     }
115 }
116 
117 void DFST(int s)
118 {
119     if(s == v1)
120     {
121         pathT.push_back(v1);
122         if(pathT.size() < optLeth)
123         {
124             optLeth = pathT.size();
125             optT = pathT;
126         }
127         pathT.pop_back();
128         return;
129     }
130     pathT.push_back(s);
131     for(int i=0; i<preT[s].size(); i++)
132         DFST(preT[s][i]);
133     pathT.pop_back();
134 }
135 
136 int main()
137 {
138 #ifdef ONLINE_JUDGE
139 #else
140     freopen("Test.txt", "r", stdin);
141 #endif // ONLINE_JUDGE
142 
143     fill(grap[0],grap[0]+M*M,INF);
144     fill(cost[0],cost[0]+M*M,INF);
145 
146     scanf("%d%d", &n,&m);
147     for(int i=0; i<m; i++)
148     {
149         scanf("%d%d%d", &v1,&v2,&one);
150         scanf("%d%d", &grap[v1][v2],&cost[v1][v2]);
151         if(!one)
152         {
153             grap[v2][v1]=grap[v1][v2];
154             cost[v2][v1]=cost[v1][v2];
155         }
156     }
157     scanf("%d%d", &v1,&v2);
158     DijskraL(v1);
159     DFSL(v2);
160     DijskraT(v1);
161     DFST(v2);
162     if(optT == optL)
163     {
164         printf("Distance = %d; Time = %d: %d", d[v2],c[v2],optT[optT.size()-1]);
165         for(int i=optT.size()-2; i>=0; i--)
166             printf(" -> %d", optT[i]);
167     }
168     else
169     {
170         printf("Distance = %d: %d", d[v2], optL[optL.size()-1]);
171         for(int i=optL.size()-2; i>=0; i--)
172             printf(" -> %d", optL[i]);
173         printf("\nTime = %d: %d", c[v2], optT[optT.size()-1]);
174         for(int i=optT.size()-2; i>=0; i--)
175             printf(" -> %d", optT[i]);
176     }
177 
178     return 0;
179 }

 

posted @ 2019-06-16 18:32  林東雨  阅读(219)  评论(0)  编辑  收藏  举报