PAT 1111. Online Map (30)

NO IMAGE

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,

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

原題


思路:
這道題其實不難,就是麻煩。利用Dijkstra演算法求最短路徑即可。
但是這裡的最短路徑分別是最短長度(長度相同時要時間短的)和最短時間(時間相同時要經過的點少的)
而要求這2個量需要使用prev陣列來儲存路徑。當遇到相同長度的值時,需要建立2條路徑來比對。因此需要注意恢復。

  int t=prev[i];
  ans=get_path(prev,i);
  prev[i]=v1;
  auto tt=get_path(prev,i);
  if(f2(tt)<f2(ans))
    ans=tt;
  else
  prev[i]=t;

正確程式碼:

#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <cmath>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <limits>
#include <assert.h>
#include <memory.h>
#include <map>
#include <stack>
using namespace std;
#define INT_MAX       2147483647
#define repeat(n) for(int _i=0;_i<n;_i  )
typedef long long ll;
int N,M;
#define INPUT cin>>N>>M
#define get_v1(v) (v/500)
#define get_v2(v) (v%500)
int road[501][501]={0};
int len[501][501]={0},u_time[501][501]={0};
int dist1[501];
int src,dst;
typedef vector<int> path;
int cal_time(const path& p)
{
int s=0;
for(int i=1;i<p.size();i  )
s =u_time[p[i-1]][p[i]];
return s;
}
int cal_path(const path& p)
{
int s=0;
for(int i=1;i<p.size();i  )
s =len[p[i-1]][p[i]];
return s;
}
path get_path(int *prev,int s)
{
path r;r.push_back(s);
while(prev[s]!=-1)
{
r.push_back(prev[s]);
s=prev[s];
}
return r;
}
int cal_intersections(const path& p){return p.size();}
void print_path(const path& p)
{
for(int i=p.size()-1;i>0;i--)
printf("%d -> ",p[i]);
cout<<p[0]<<endl;
}
typedef int (* fun)(const path& );
path find_path(int* dist,int len[][501],fun f1,fun f2)
{
dist[src]=0;
int visited[501]={0};
int prev[501]={0};
memset(prev,-1,501*4);
path ans;
while(1)
{
int v1=-1;
for(int i=0;i<N;i  )
if(!visited[i]){
if(v1==-1) v1=i; else
if(dist[i]<dist[v1])
v1=i;
}
//find min v1 
visited[v1]=1;
if(v1==dst)
{
return get_path(prev,dst);
}
if(dist[v1]>dist[dst])
assert(0);
for(int i=0;i<N;i  )
if(road[v1][i]&&(!visited[i]))
{
if(dist[i]>dist[v1] len[v1][i]){
dist[i]=dist[v1] len[v1][i];
prev[i]=v1;
}else if(dist[i]==dist[v1] len[v1][i])
{
int t=prev[i];
ans=get_path(prev,i);
prev[i]=v1;
auto tt=get_path(prev,i);
if(f2(tt)<f2(ans))
ans=tt;
else
prev[i]=t;
}
}
}
return ans;
}
int use_time[501];
int main()
{
#ifdef _DEBUG
fstream cin("input.txt");
#endif
(INPUT);
for(int i=0;i<M;i  )
{
int v1,v2,one,l,t;
cin>>v1>>v2>>one>>l>>t;
road[v1][v2]=1;
if(!one)
road[v2][v1]=1;
len[v1][v2]=len[v2][v1]=l;
u_time[v1][v2]=u_time[v2][v1]=t;
assert(v1<N&&v2<N);
}
cin>>src>>dst;
assert(N<501);
//end input
for(int i=0;i<N;i  ) dist1[i]=INT_MAX;
const auto & p1=find_path(dist1,len,&cal_path,&cal_time);
for(int i=0;i<N;i  ) dist1[i]=INT_MAX;
const auto & p2=find_path(dist1,u_time,&cal_time,&cal_intersections );
if(p1==p2)
{
printf("Distance = %d; Time = %d: ",cal_path(p1),cal_time(p1));
print_path(p1);
}
else{
printf("Distance = %d: ",cal_path(p1));
print_path(p1);
printf("Time = %d: ",cal_time(p2));
print_path(p2);
}
return 0;
}

使用vector來儲存路徑,用set來取得最小路徑。結構更容易理解。
不知道為什麼有一個點出問題:

#include <cstdio>
#include <string>
#include <vector>
#include <set>
#include <cmath>
#include <algorithm>
#include <fstream>
#include <iostream>
#include <limits>
#include <assert.h>
#include <map>
#include <stack>
using namespace std;
#define INT_MAX       2147483647
#define repeat(n) for(int _i=0;_i<n;_i  )
typedef long long ll;
int N,M;
#define INPUT cin>>N>>M
#define get_v1(v) (v/500)
#define get_v2(v) (v%500)
int road[501][501]={0};
int len[501][501]={0},u_time[501][501]={0};
int dist1[501];
int src,dst;
typedef vector<int> path;
int cal_time(const path& p)
{
int s=0;
for(int i=1;i<p.size();i  )
s =u_time[p[i-1]][p[i]];
return s;
}
int cal_path(const path& p)
{
int s=0;
for(int i=1;i<p.size();i  )
s =len[p[i-1]][p[i]];
return s;
}
int cal_intersections(const path& p){return p.size();}
void print_path(const path& p)
{
for(int i=0;i<p.size()-1;i  )
printf("%d -> ",p[i]);
cout<<p[p.size()-1]<<endl;
}
typedef int (* fun)(const path& );
path find_path(int* dist,int len[][501],fun f1,fun f2)
{
auto f=[&f1,&f2](const path& v1,const path& v2)->bool
{
int l1=f1(v1),l2=f1(v2);
if(l1==l2)
return f2(v1)<f2(v2);
return l1<l2;
};
set<path,decltype(f)> search(f);
dist[src]=0;
int visited[501]={0};
int min_len=INT_MAX;
path ans(1,src);
search.insert(ans);
while(!search.empty())
{
const auto p1=*search.begin();
int v1=*p1.rbegin();
visited[v1]=1;
search.erase(search.begin());
if(v1==dst)
{
return p1;
}
if(dist[v1]>dist[dst])
break;
for(int i=0;i<N;i  )
if(road[v1][i]&&(dist[i]>=dist[v1] len[v1][i]))
{
dist[i]=dist[v1] len[v1][i];
auto t=p1;
t.push_back(i);
search.insert(t);
}
}
while(1) dist;
assert(0);
return ans;
}
int use_time[501];
int main()
{
#ifdef _DEBUG
fstream cin("input.txt");
#endif
(INPUT);
for(int i=0;i<M;i  )
{
int v1,v2,one,l,t;
cin>>v1>>v2>>one>>l>>t;
road[v1][v2]=1;
if(!one)
road[v2][v1]=1;
len[v1][v2]=len[v2][v1]=l;
u_time[v1][v2]=u_time[v2][v1]=t;
assert(v1<N&&v2<N);
}
cin>>src>>dst;
assert(N<501);
//end input
for(int i=0;i<N;i  ) dist1[i]=INT_MAX;
const auto & p1=find_path(dist1,len,&cal_path,&cal_time);
for(int i=0;i<N;i  ) dist1[i]=INT_MAX;
const auto & p2=find_path(dist1,u_time,&cal_time,&cal_intersections );
if(p1==p2)
{
printf("Distance = %d; Time = %d: ",cal_path(p1),cal_time(p1));
print_path(p1);
}
else{
printf("Distance = %d: ",cal_path(p1));
print_path(p1);
printf("Time = %d: ",cal_time(p2));
print_path(p2);
}
return 0;
}