这道题我读题不仔细导致踩了个大坑，一个测试点过不了卡了好几个小时：第二个dijkstra算法中，题目要求是“In case the fastest path is not unique, output the one that passes through the fewest intersections”，我却想当然地认为在fastest path is not unique的时候，判断标准是最短距离……
#include <cstdio>
#include <vector>
#include <algorithm>
const int MAXN = 501;
const int INF = 999999999;
int N, M, u, v, oneWay, a, b, src, dst, tmp;
int dist[MAXN][MAXN];
int time[MAXN][MAXN];
int path[MAXN];
std::vector<int> d1, t1, t2, path1, path2;
void dijkstra1(int k){
int minDist, pivot;
d1.resize(N, INF);
t1.resize(N, INF);
std::vector<bool> visited(N);
std::fill(visited.begin(), visited.end(), false);
std::fill(path, path + N, -1);
d1[k] = 0;
t1[k] = 0;
for(int i = 0; i < N; ++i){
pivot = -1;
minDist = INF;
for(int j = 0; j < N; ++j){
if(!visited[j] && d1[j] < minDist){
pivot = j;
minDist = d1[j];
}
}
if(pivot == -1){
return;
}
visited[pivot] = true;
for(int j = 0; j < N; ++j){
if(!visited[j] && dist[pivot][j] < INF && d1[pivot] + dist[pivot][j] < d1[j]){
d1[j] = d1[pivot] + dist[pivot][j];
t1[j] = t1[pivot] + time[pivot][j];
path[j] = pivot;
} else if(!visited[j] && dist[pivot][j] < INF && time[pivot][j] < INF
&& d1[pivot] + dist[pivot][j] == d1[j] && t1[pivot] + time[pivot][j] < t1[j]){
t1[j] = t1[pivot] + time[pivot][j];
path[j] = pivot;
}
}
}
}
void dijkstra2(int k){
int minTime, pivot;
t2.resize(N, INF);
std::vector<int> intersection(N, INF);
std::vector<bool> visited(N);
std::fill(visited.begin(), visited.end(), false);
std::fill(path, path + N, -1);
intersection[k] = 0;
t2[k] = 0;
for(int i = 0; i < N; ++i){
pivot = -1;
minTime = INF;
for(int j = 0; j < N; ++j){
if(!visited[j] && t2[j] < minTime){
pivot = j;
minTime = t2[j];
}
}
if(pivot == -1){
return;
}
visited[pivot] = true;
for(int j = 0; j < N; ++j){
if(!visited[j] && time[pivot][j] < INF && t2[pivot] + time[pivot][j] < t2[j]){
t2[j] = t2[pivot] + time[pivot][j];
intersection[j] = intersection[pivot] + 1;
path[j] = pivot;
} else if(!visited[j] && time[pivot][j] < INF
&& t2[pivot] + time[pivot][j] == t2[j] && intersection[pivot] + 1 < intersection[j]){
intersection[j] = intersection[pivot] + 1;
path[j] = pivot;
}
}
}
}
int main(){
std::fill(dist[0], dist[0] + MAXN * MAXN, INF);
std::fill(time[0], time[0] + MAXN * MAXN, INF);
scanf("%d %d", &N, &M);
for(int i = 0; i < M; ++i){
scanf("%d %d %d %d %d", &u, &v, &oneWay, &a, &b);
dist[u][v] = a;
time[u][v] = b;
if(!oneWay){
dist[v][u] = a;
time[v][u] = b;
}
}
scanf("%d %d", &src, &dst);
dijkstra1(src);
tmp = dst;
while(path[tmp] != -1){
path1.push_back(tmp);
tmp = path[tmp];
}
dijkstra2(src);
tmp = dst;
while(path[tmp] != -1){
path2.push_back(tmp);
tmp = path[tmp];
}
if(path1 == path2){
printf("Distance = %d; Time = %d: %d", d1[dst], t1[dst], src);
for(int i = path1.size() - 1; i >= 0; --i){
printf(" -> %d", path1[i]);
}
} else{
printf("Distance = %d: %d", d1[dst], src);
for(int i = path1.size() - 1; i >= 0; --i){
printf(" -> %d", path1[i]);
}
printf("\nTime = %d: %d", t2[dst], src);
for(int i = path2.size() - 1; i >= 0; --i){
printf(" -> %d", path2[i]);
}
}
return 0;
}题目如下：
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.文章来源：https://www.toymoban.com/news/detail-468927.html
In case the shortest and the fastest paths are identical, print them in one line in the format:文章来源地址https://www.toymoban.com/news/detail-468927.html
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到了这里，关于1111 Online Map （PAT甲级）的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
