1111. Online Map (30)-PAT甲级真题（Dijkstra + DFS）

1111.
Online Map (30)
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：用两个Dijkstra
+ DFS。一个求最短路径（如果相同求时间最短的那条），一个求最快路径（如果相同求结点数最小的那条）~~~求最短路径可以直接在Dijkstra里面求前驱结点pre数组~~~求最快路径因为要求结点数最小的那条，所以要用pre的二维数组存储所有结点的最快路径，然后用DFS求出满足条件的结点数最小的那条~~~~
注意：
1.一开始最后一个测试用例“答案错误”，后来发现是自己在求最快路径的时候忘记了temppath每一次深搜结束后的pop_back();

2.如果直接使用DFS的话，会导致最后一个测试用例“运行超时”~~~

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int inf = 999999999;
int dis[510], Time[510], e[510][510], w[510][510], dispre[510], weight[510];
bool visit[510];
vector<int> dispath, Timepath, temppath, Timepre[510];
int st, fin, minnode = inf;
void dfsdispath(int v) {
dispath.push_back(v);
if(v == st) {
return ;
}
dfsdispath(dispre[v]);
}

void dfsTimepath(int v) {
temppath.push_back(v);
if(v == st) {
if(temppath.size() < minnode) {
minnode = temppath.size();
Timepath = temppath;
}
temppath.pop_back();
return ;
}
for(int i = 0; i < Timepre[v].size(); i++) {
dfsTimepath(Timepre[v][i]);
}
temppath.pop_back();
}
int main() {
fill(dis, dis + 510, inf);
fill(Time, Time + 510, inf);
fill(weight, weight + 510, inf);
fill(e[0], e[0] + 510 * 510, inf);
fill(w[0], w[0] + 510 * 510, inf);
int n, m;
scanf("%d %d", &n, &m);
int a, b, flag, len, t;
for(int i = 0; i < m; i++) {
scanf("%d %d %d %d %d", &a, &b, &flag, &len, &t);
e[a][b] = len;
w[a][b] = t;
if(flag != 1) {
e[b][a] = len;
w[b][a] = t;
}
}
scanf("%d %d", &st, &fin);
dis[st] = 0;
for(int i = 0; i < n; i++) {
dispre[i] = i;
}
for(int i = 0; i < n; i++) {
int u = -1, minn = inf;
for(int j = 0; j < n; j++) {
if(visit[j] == false && dis[j] < minn) {
u = j;
minn = dis[j];
}
}
if(u == -1) break;
visit[u] = true;
for(int v = 0; v < n; v++) {
if(visit[v] == false && e[u][v] != inf) {
if(e[u][v] + dis[u] < dis[v]) {
dis[v] = e[u][v] + dis[u];
dispre[v] = u;
weight[v] = weight[u] + w[u][v];
} else if(e[u][v] + dis[u] == dis[v] && weight[v] > weight[u] + w[u][v]) {
weight[v] = weight[u] + w[u][v];
dispre[v] = u;
}
}
}
}
dfsdispath(fin);
Time[st] = 0;
fill(visit, visit + 510, false);
for(int i = 0; i < n; i++) {
int u = -1, minn = inf;
for(int j = 0; j < n; j++) {
if(visit[j] == false && minn > Time[j]) {
u = j;
minn = Time[j];
}
}
if(u == -1) break;
visit[u] = true;
for(int v = 0; v < n; v++) {
if(visit[v] == false && w[u][v] != inf) {
if(w[u][v] + Time[u] < Time[v]) {
Time[v] = w[u][v] + Time[u];
Timepre[v].clear();
Timepre[v].push_back(u);
} else if(w[u][v] + Time[u] == Time[v]) {
Timepre[v].push_back(u);
}
}
}
}
dfsTimepath(fin);
printf("Distance = %d", dis[fin]);
if(dispath == Timepath) {
printf("; Time = %d: ", Time[fin]);
} else {
printf(": ");
for(int i = dispath.size() - 1; i >= 0; i--) {
printf("%d", dispath[i]);
if(i != 0) printf(" -> ");
}
printf("\n");
printf("Time = %d: ", Time[fin]);
}
for(int i = Timepath.size() - 1; i >= 0; i--) {
printf("%d", Timepath[i]);
if(i != 0) printf(" -> ");
}
return 0;
}