)
思路
dijkstra模板题，不需要小根堆优化，这题的第二个样例：路径一样则合并输出，怎么判断路径一样？会想到C++的vector对==进行了重写（即每个对应位置的元素一样）。
AcCode：
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
int N, M;
//最后一维表示time或distance，graph存图，dist存最短距，lst存上个节点
int graph[510][510][2], dist[510][2], lst[510][2];
//两种路径
vector<int> path[2];
//dijkstra模板
int dj(int st, int be, int end){dist[be][st] = 0;bool vis[510];lst[be][st] = -1;memset(vis, false, sizeof(vis));for(int i = 0; i < N; i++){int cur, mn = 0x7fffffff;for(int i = 0; i < N; i++) if(dist[i][st] < mn && !vis[i]) mn = dist[i][st], cur = i; for(int i = 0; i < N; i++){if(graph[cur][i][st] && dist[cur][st] + graph[cur][i][st] < dist[i][st] && !vis[i]){dist[i][st] = dist[cur][st] + graph[cur][i][st];lst[i][st] = cur;}}vis[cur] = true;}// for(int i = 0; i < N; i++) cout << lst[i][st] << "-";// cout << endl;return dist[end][st];
}
//递归的获取路径
void getPath(int st, int end){if(end == -1) return;getPath(st, lst[end][st]);path[st].push_back(end);
}int main(){
//初始化dist数组为大值，并不知道time和lenth取值范围，题目压根没写，我猜的。memset(dist, 0x2f, sizeof(dist));cin >> N >> M;
//输入数据while(M--){int v1, v2, ow, len, time; cin >> v1 >> v2 >> ow >> len >> time;graph[v1][v2][0] = time;graph[v1][v2][1] = len;if(ow != 1) graph[v2][v1][0] = time, graph[v2][v1][1] = len;}
//调用函数输出答案即可int be, end; cin >> be >> end;int ans[2];for(int i = 0; i < 2; i++){ans[i] = dj(i, be, end);getPath(i, end);}if(path[1] != path[0]){cout << "Time = " << ans[0] << ": ";for(int i = 0; i < path[0].size(); i++){cout << path[0][i];if(i != path[0].size() - 1) cout << " => ";}cout << endl << "Distance = " << ans[1] << ": ";for(int i = 0; i < path[1].size(); i++){cout << path[1][i];if(i != path[1].size() - 1) cout << " => ";}}else{cout << "Time = " << ans[0] << "; " << "Distance = " << ans[1] << ": ";for(int i = 0; i < path[0].size(); i++){cout << path[0][i];if(i != path[0].size() - 1) cout << " => ";}}return 0;
}