pat--1111. Online Map

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:

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 = 8, 4, 6:3, 5
Time = 5 3:3 , 1, 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 = 2--5 4:3

Shortest path, each edge has two weights, distance and time.
Taking distance as the measure, the shortest path with shortest distance is obtained.
Taking time as a measure, we find the shortest path with the shortest time through the least node.

#include <iostream> #include <cstdio> #include <algorithm> #include <vector> #include <

Cstring> using namespace std;
const int MAXN = 500 + 10;
const int inf = 0X3F3F3F3F;
int N, M, S, E;
int _DIS[MAXN][MAXN], _TIME[MAXN][MAXN];
int DIS[MAXN], TIME_DIS[MAXN], DISTIME[MAXN], CNT[MAXN];
int VIS[MAXN];
int DISPATH[MAXN], TIMEPATH[MAXN];

Vector<int> Ans_dis_path, Ans_time_path;
    void Distancedij () {memset (Vis, 0, sizeof (VIS));
    for (int i = 0; i < N; ++i) dis[i] = (i = = S? 0:inf);
        for (int i = 0; i < N; ++i) {int x, m = inf;
        for (int y = 0; y < N; ++y) if (!vis[y] && dis[y] <= m) m = dis[x=y];
        VIS[X] = 1; for (int y = 0; y < N; ++y) {if (!vis[y] && _dis[x][y]) {if (Dis[y] > dis[x] + _dis [x] [y])
                    {Dis[y] = Dis[x] + _dis[x][y];
                    Dispath[y] = x;
               Distime[y] = Distime[x] + _time[x][y]; }else if (dis[y] = = Dis[x] + _dis[x][y] && distime[y] > Distime[x] + _time[x][y]) {distime[
                    Y] = distime[x] + _time[x][y];
                Dispath[y] = x;
    }}}}} void Timedij () {memset (Vis, 0, sizeof (VIS));
    for (int i = 0; i < N; ++i) time_dis[i] = (i = = S? 0:inf);
    Vis[s] = 1;
        for (int i = 0; i < N; ++i) {int x, m = inf;
        for (int y = 0; y < N; ++y) if (!vis[y] && time_dis[y] <= m) m = time_dis[x=y];
        VIS[X] = 1; for (int y = 0; y < N; ++y) {if (!vis[y] && _time[x][y]) {if (Time_dis[y] > Time_di
                    S[X] + _time[x][y]) {time_dis[y] = Time_dis[x] + _time[x][y];
                    Timepath[y] = x;
                Cnt[y] = cnt[x] + 1; 
                 }else if (time_dis[y] = = Time_dis[x] + _time[x][y] && cnt[y] > Cnt[x] + 1) {timepath[y] = x;   Cnt[y] = cnt[x] + 1;
    }}}}} void Getdispath () {int tmp = E;
        while (tmp! = S) {ans_dis_path.push_back (TMP);
    TMP = dispath[tmp];
} ans_dis_path.push_back (S);
    } void Gettimepath () {int tmp = E;
        while (tmp! = S) {ans_time_path.push_back (TMP);
    TMP = timepath[tmp];
} ans_time_path.push_back (S);
    } int main () {scanf ("%d%d", &n, &m);
        for (int i = 0; i < M; ++i) {int u, V, O, L, T;
        scanf ("%d%d%d%d", &u, &v, &o, &l, &t);
        _DIS[U][V] = l, _time[u][v] = t;
    if (!o) _dis[v][u] = l, _time[v][u] = t;

    } scanf ("%d%d", &s, &e);    
    Distancedij ();

    Timedij ();
    Getdispath ();

    Gettimepath (); if (Ans_dis_path = = Ans_time_path) {printf ("Distance =%d;
        Time =%d: ", Dis[e], time_dis[e]);
        int i = 0; for (Auto it = Ans_dis_path.rbegin (); It! = Ans_dis_path.rend (); ++it) {
            if (i++) printf ("-");
        printf ("%d", *it);
        }}else{printf ("Distance =%d:", dis[e]);
        int i = 0;
            for (Auto it = Ans_dis_path.rbegin (); It! = Ans_dis_path.rend (); ++it) {if (i++) printf ("-");
        printf ("%d", *it);
        } printf ("\ntime =%d:", time_dis[e]);
        i = 0;
            for (Auto it = Ans_time_path.rbegin (); It! = Ans_time_path.rend (); ++it) {if (i++) printf ("-");
        printf ("%d", *it);
}} return 0;
 }