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
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分析：Dijkstra模板题，套2次就行了。
1 /**
2 * Copyright(c)
3 * All rights reserved.
4 * Author : Mered1th
5 * Date : 2019-02-26-22.31.18
6 * Description : A1111
7 */
8 #include<cstdio>
9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=510;
20 const int INF=1000000000;
21 bool vis[maxn];
22 int G[maxn][maxn],d[maxn],t[maxn],T[maxn][maxn];
23 int n,m,st,ed;
24 int pred[maxn],pret[maxn],mininter[maxn];
25 vector<int> shortPath,fastPath;
26 void dDijkstra(int s){
27 fill(d,d+maxn,INF);
28 fill(t,t+maxn,INF);
29 d[s]=0;
30 t[s]=0;
31 memset(vis,0,sizeof(vis));
32 for(int i=0;i<n;i++) pred[i]=i;
33 for(int i=0;i<n;i++){
34 int u=-1,MIN=INF;
35 for(int j=0;j<n;j++){
36 if(vis[j]==false && d[j]<MIN){
37 u=j;
38 MIN=d[j];
39 }
40 }
41 if(u==-1) return;
42 vis[u]=true;
43 for(int v=0;v<n;v++){
44 if(vis[v]==false && G[u][v]!=INF){
45 if(G[u][v]+d[u]<d[v]){
46 d[v]=d[u]+G[u][v];
47 t[v]=t[u]+T[u][v];
48 pred[v]=u;
49 }
50 else if(G[u][v]+d[u]==d[v]){
51 if(T[u][v]+t[u]<t[v]){
52 pred[v]=u;
53 t[v]=T[u][v]+t[u];
54 }
55 }
56 }
57 }
58 }
59 for(int i=ed;i!=st;i=pred[i]){
60 shortPath.push_back(i);
61 }
62 shortPath.push_back(st);
63 }
64
65 void tDijkstra(int s){
66 fill(t,t+maxn,INF);
67 t[s]=0;
68 fill(mininter,mininter+maxn,1);
69 memset(vis,0,sizeof(vis));
70 for(int i=0;i<n;i++) pret[i]=i;
71 for(int i=0;i<n;i++){
72 int u=-1,MIN=INF;
73 for(int j=0;j<n;j++){
74 if(vis[j]==false && t[j]<MIN){
75 u=j;
76 MIN=t[j];
77 }
78 }
79 if(u==-1) return;
80 vis[u]=true;
81 for(int v=0;v<n;v++){
82 if(vis[v]==false && T[u][v]!=INF){
83 if(T[u][v]+t[u]<t[v]){
84 t[v]=t[u]+T[u][v];
85 pret[v]=u;
86 mininter[v]=mininter[u]+1;
87 }
88 else if(T[u][v]+t[u]==t[v]){
89 if(mininter[v]>mininter[u]+1){
90 pret[v]=u;
91 mininter[v]=mininter[u]+1;
92 }
93 }
94 }
95 }
96 }
97 for(int i=ed;i!=st;i=pret[i]){
98 fastPath.push_back(i);
99 }
100 fastPath.push_back(st);
101 }
102
103
104 int main(){
105 #ifdef ONLINE_JUDGE
106 #else
107 freopen("1.txt", "r", stdin);
108 #endif
109 int u,v,o;
110 cin>>n>>m;
111 fill(G[0],G[0]+maxn*maxn,INF);
112 fill(T[0],T[0]+maxn*maxn,INF);
113 for(int i=0;i<m;i++){
114 cin>>u>>v>>o;
115 cin>>G[u][v]>>T[u][v];
116 if(o==0){
117 G[v][u]=G[u][v];
118 T[v][u]=T[u][v];
119 }
120 }
121 cin>>st>>ed;
122 dDijkstra(st);
123 tDijkstra(st);
124 if(shortPath!=fastPath){
125 printf("Distance = %d: ",d[ed]);
126 for(int i=shortPath.size()-1;i>=0;i--){
127 printf("%d",shortPath[i]);
128 if(i!=0) printf(" -> ");
129 else printf("\n");
130 }
131 printf("Time = %d: ",t[ed]);
132 for(int i=fastPath.size()-1;i>=0;i--){
133 printf("%d",fastPath[i]);
134 if(i!=0) printf(" -> ");
135 else printf("\n");
136 }
137 }
138 else{
139 printf("Distance = %d; Time = %d: ",d[ed],t[ed]);
140 for(int i=fastPath.size()-1;i>=0;i--){
141 printf("%d",fastPath[i]);
142 if(i!=0) printf(" -> ");
143 else printf("\n");
144 }
145 }
146 return 0;
147 }