PAT advanced 1111. Online Map (30)

Problem Description:

1111. Online Map ( time limit)
Memory Limit 65536 KB
Code length limit 16000 B
Program Standard author CHEN, Yue

Input position and a destination, an online map can recommend several paths. Now your job are to recommend, paths to your user:one are the shortest, and the other is the fastest. It is guaranteed, a path exists for any request.

Input Specification:

Each input file contains the one test case. For each case, the first line gives-positive integers n (2 <= n <=), and M, being the total number of the street s 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 is the indices (from 0 to N-1) of the "The" of the street; one-way is 1 if the street was 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 on the next line print the fastest path with total time T:

Time = T:source, W1---destination

In case the shortest path isn't unique, output the fastest one among the shortest paths, which is guaranteed to be unique . In case the fastest path isn't unique, output the one that passes through the fewest intersections, which is guaranteed t o be unique.

In case the shortest and the fastest paths is identical, print them in one line in the format:

Distance = D; Time = T:source-u1-Destination Sample Input 1:

Ten
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 = 8, 4, 6:3, 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 = 5 4:3, 2

I began to review the problem in advanced ...

There is no short-circuit problem can not be solved with heap optimization Dijkstra algorithm, if there is, use two times ...

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#include <bits/stdc++.h> using namespace std; struct node {int id; int Len; int Tim; int jump; int pid; } No; struct CMP1 {bool operator () (const node & e1,const Node & E2) Const {if (E1.len!=e2.len) return E1.len&gt ; E2.len; else return e1.tim>e2.tim; } }; struct CMP2 {bool operator () (const node & e1,const Node & E2) Const {if (E1.tim!=e2.tim) return E1.tim&gt ; e2.tim; else return e1.jump>e2.jump; } }; vector<vector<node> > V; int main () {//Ios::sync_with_stdio (FALSE);//Freopen ("Data.txt", "R", stdin); int n,m,k,c1,c2,x,t,l,e11,e12,e13,e21,e22,e23; scanf ("%d%d", &n,&m); V.resize (n); vector<int> vb (n,-1); No.pid=0; No.jump=0; for (; m--;) {scanf ("%d%d%d%d%d", &c1,&c2,&k,&no.len,&no.tim); NO.ID=C2; V[c1].push_back (no); if (!k) {no.id=c1; V[c2].push_back (no); }} scanf ("%d%d", &AMP;C1,&AMP;C2); Stack<int> s1,s2; NO.ID=C1; No.jump=0; no.len=0; No.tim=0; NO.PID=C1; Priority_queue<node,vector<node>,cmp1> Q1; Q1.push (no); for (;! Q1.empty ();) {no=q1.top (); X=no.id; L=no.len; T=no.tim; K=no.jump; if (vb[x]<0) {vb[x]=no.pid; if (X==C2) {E11=no.len; E12=no.tim; E13=no.jump; Break } for (int i=0;i<v[x].size (); i++) {no.pid=x; No.id=v[x][i].id; No.len=l+v[x][i].len; No.tim=t+v[x][i].tim; no.jump=k+1; Q1.push (no); }} q1.pop (); } for (int i=c2;i!=c1;i=vb[i]) S1.push (i); Vb.clear (); Vb.resize (n,-1); NO.ID=C1; No.jump=0; no.len=0; no.tim=0; NO.PID=C1; Priority_queue<node,vector<node>,cmp2> Q2; Q2.push (no); for (;! Q2.empty ();) {no=q2.top (); X=no.id; L=no.len; T=no.tim; K=no.jump; if (vb[x]<0) {vb[x]=no.pid; if (X==C2) {E21=no.len; E22=no.tim; E23=no.jump; Break } for (int i=0;i<v[x].size (); i++) {no.pid=x; No.id=v[x][i].id; No.len=l+v[x][i].len; No.tim=t+v[x][i].tim; no.jump=k+1; Q2.push (no); }} q2.pop (); } for (int i=c2;i!=c1;i=vb[i]) S2.push (i); if (e11==e21&&e12==e22&&e13==e23) {printf ("Distance =%d; Time =%d: ", e11,e12); printf ("%d", C1); for (;! S1.empty (); S1.pop ()) printf ("-%d", s1.top ()); } else {printf ("Distance =%d:", E11); printf ("%d", C1); for (;! S1.empty (); S1.pop ()) printf ("-%d", s1.top ()); printf ("\ n"); printf ("Time =%d:", E22); printf ("%d", C1); for (;! S2.empty (); S2.pop ()) printf ("-%d", s2.top ()); } return 0; }