一、题目
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
二、题目大意
求最短路径和最时间。
三、考点
DFS、Dijkstra
四、注意
1、单独使用DFS会有一个测试点超时;
2、该代码单独使用DFS;
3、Dijkstra比DFS节省时间。
五、代码
#include<iostream>
#include<string>
#include<map>
#include<vector>
#define N 510
#define INF 9999999
using namespace std;
int mstart=1, mend;
vector<int> vec_happy;
int e[N][N],t[N][N];
bool visit[N] = { false };
int out_dis=INF, out_time=INF, out_dis_time=INF;
vector<int> vec_out_dis,vec_out_time, vec_path;
int n, m;
void dfs(int root, int dis, int time) {
//return
if (dis > out_dis && time>out_time)
return;
//arrive at mend
if (root == mend) {
//dis
if (dis < out_dis || dis == out_dis && time<out_dis_time) {
out_dis = dis;
vec_out_dis = vec_path;
out_dis_time = time;
}
//time
if (time < out_time || vec_path.size() <vec_out_time.size() && time==out_time) {
out_time = time;
vec_out_time = vec_path;
}
return;
}
//next dfs
for (int i = 0; i < n; ++i) {
if (visit[i] == false && e[root][i] != INF) {
visit[i] = true;
vec_path.push_back(i);
dfs(i, dis + e[root][i], time + t[root][i]);
visit[i] = false;
vec_path.pop_back();
}
}
return;
}
int main() {
//init
fill(e[0], e[0] + N*N, INF);
fill(t[0], t[0] + N*N, INF);
fill(visit, visit + N, false);
//read
cin >> n >> m;
for (int i = 0; i < m; ++i) {
int a, b, c, d, f;
//cin >> a >> b >> c >> d >> f;
scanf("%d %d %d %d %d", &a, &b, &c, &d, &f);
if (c == 1) {
e[a][b] = d;
t[a][b] = f;
}
else {
e[a][b] =e[b][a]= d;
t[a][b] =t[b][a]= f;
}
}
cin >> mstart >> mend;
//dfs
vec_path.push_back(mstart);
visit[mstart] = true;
dfs(mstart, 0, 0);
//output
if (vec_out_dis != vec_out_time) {
printf("Distance = %d: ", out_dis);
for (int i = 0; i < vec_out_dis.size(); ++i) {
if (i != 0)
cout << " -> ";
cout << vec_out_dis[i];
}
printf("\nTime = %d: ", out_time);
for (int i = 0; i < vec_out_time.size(); ++i) {
if (i != 0)
cout << " -> ";
cout << vec_out_time[i];
}
}
else {
printf("Distance = %d; Time = %d: ", out_dis, out_time);
for (int i = 0; i < vec_out_dis.size(); ++i) {
if (i != 0)
cout << " -> ";
cout << vec_out_dis[i];
}
}
cout << endl;
system("pause");
return 0;
}