L3-007 天梯地图 (30 分)
本题要求你实现一个天梯赛专属在线地图,队员输入自己学校所在地和赛场地点后,该地图应该推荐两条路线:一条是最快到达路线;一条是最短距离的路线。题目保证对任意的查询请求,地图上都至少存在一条可达路线。
输入格式:
输入在第一行给出两个正整数N(2 ≤ N ≤ 500)和M,分别为地图中所有标记地点的个数和连接地点的道路条数。随后M行,每行按如下格式给出一条道路的信息:
V1 V2 one-way length time
其中V1和V2是道路的两个端点的编号(从0到N-1);如果该道路是从V1到V2的单行线,则one-way为1,否则为0;length是道路的长度;time是通过该路所需要的时间。最后给出一对起点和终点的编号。
输出格式:
首先按下列格式输出最快到达的时间T和用节点编号表示的路线:
Time = T: 起点 => 节点1 => … => 终点
然后在下一行按下列格式输出最短距离D和用节点编号表示的路线:
Distance = D: 起点 => 节点1 => … => 终点
如果最快到达路线不唯一,则输出几条最快路线中最短的那条,题目保证这条路线是唯一的。而如果最短距离的路线不唯一,则输出途径节点数最少的那条,题目保证这条路线是唯一的。
如果这两条路线是完全一样的,则按下列格式输出:
Time = T; Distance = D: 起点 => 节点1 => … => 终点
输入样例1:
10 15
0 1 0 1 1
8 0 0 1 1
4 8 1 1 1
5 4 0 2 3
5 9 1 1 4
0 6 0 1 1
7 3 1 1 2
8 3 1 1 2
2 5 0 2 2
2 1 1 1 1
1 5 0 1 3
1 4 0 1 1
9 7 1 1 3
3 1 0 2 5
6 3 1 2 1
5 3
输出样例1:
Time = 6: 5 => 4 => 8 => 3
Distance = 3: 5 => 1 => 3
输入样例2:
7 9
0 4 1 1 1
1 6 1 3 1
2 6 1 1 1
2 5 1 2 2
3 0 0 1 1
3 1 1 3 1
3 2 1 2 1
4 5 0 2 2
6 5 1 2 1
3 5
输出样例2:
Time = 3; Distance = 4: 3 => 2 => 5
题解
写两份不同度量下的最短路算法,对于相等情况做特殊处理。
代码
#include <bits/stdc++.h>
using namespace std;
const static int INF = 0x3fffffff;
struct Edge
{
int val;
int t;
Edge(int _t, int _v)
{
t = _t;
val = _v;
}
bool operator<(const Edge &a) const
{
return val > a.val;
}
};
int Dijkstra1(vector<vector<int> > &G, int s, int t, vector<int> &path) //最短路径,节点最少
{
int n = G.size();
vector<int> dist(n,INF);
for(int i=0;i<n;i++)
path[i] = -1;
path[s] = 0;
dist[s] = 0;
vector<int> cnt(n, INF);
cnt[s] = 1;
vector<int> visit(n, false);
priority_queue<Edge> pq;
pq.push(Edge(s, dist[s]));
while(!pq.empty())
{
Edge edge = pq.top();pq.pop();
int u = edge.t;
if(u == t)
break;
if(visit[u])
continue;
visit[u] = true;
for(int v=0;v<n;v++)
{
if(G[u][v] != 0 && visit[v] == false)
{
int temp = dist[u] + G[u][v];
if(temp < dist[v])
{
dist[v] = temp;
path[v] = u;
cnt[v] = cnt[u] + 1;
pq.push(Edge(v, dist[v]));
}
else if(temp == dist[v]) //长度相同,节点最少
{
if(cnt[v] > cnt[u] + 1)
{
path[v] = u;
cnt[v] = cnt[u] + 1;
pq.push(Edge(v, dist[v]));
}
}
}
}
}
return dist[t];
}
int Dijkstra2(vector<vector<int> > &G, vector<vector<int> > &D, int s, int t, vector<int> &path) //最快路线中最短的
{
int n = G.size();
vector<int> dist(n, INF);
vector<int> cost(n, INF);
for(int i=0;i<n;i++)
path[i] = -1;
path[s] = 0;
dist[s] = cost[s] = 0;
vector<int> visit(n, false);
priority_queue<Edge> pq;
pq.push(Edge(s, cost[s]));
while(!pq.empty())
{
Edge edge = pq.top();pq.pop();
int u = edge.t;
if(u == t)
break;
if(visit[u])
continue;
visit[u] = true;
for(int v=0;v<n;v++)
{
if(G[u][v] != 0 && visit[v] == false)
{
int temp = cost[u] + G[u][v];
if(temp < cost[v])
{
cost[v] = temp;
dist[v] = dist[u] + D[u][v];
path[v] = u;
pq.push(Edge(v, cost[v]));
}
else if(temp == cost[v]) //时间相同,距离优先
{
if(dist[v] > dist[u] + D[u][v])
{
dist[v] = dist[u] + D[u][v];
path[v] = u;
pq.push(Edge(v, cost[v]));
}
}
}
}
}
return cost[t];
}
bool is_same_path(vector<int> &p1, vector<int> &p2)
{
if(p1.size() != p2.size())
return false;
for(size_t i=0;i<p1.size();i++)
{
if(p1[i] != p2[i])
return false;
}
return true;
}
vector<int> getPath(int s, int t, vector<int> &path)
{
int last = t;
vector<int> ans;
while(s != last )
{
ans.push_back(last);
last = path[last];
}
ans.push_back(s);
return ans;
}
void print(vector<int> P)
{
int n = P.size();
for(int i=n-1;i>=0;i--)
{
if(i != n-1)
cout << " => ";
cout << P[i];
}
cout << endl;
}
int main()
{
int n, m;
cin >> n >> m;
vector<vector<int> > T(n,vector<int>(n,0));
vector<vector<int> > D(n,vector<int>(n,0));
int s,t,type, len, time;
for(int i=0;i<m;i++)
{
cin >> s >> t >> type >> len >> time;
T[s][t] = time;
D[s][t] = len;
if(type == 0)
{
T[t][s] = time;
D[t][s] = len;
}
}
vector<int> tpath(n);
vector<int> dpath(n);
cin >> s >> t;
int cd = Dijkstra1(D, s, t, dpath);
int ct = Dijkstra2(T, D, s, t, tpath);
vector<int> tp = getPath(s, t, tpath);
vector<int> dp = getPath(s, t, dpath);
if(is_same_path(tp, dp))
{
cout << "Time = " << ct << "; Distance = " << cd << ": ";
print(tp);
}
else
{
cout << "Time = " << ct << ": ";
print(tp);
cout << "Distance = " << cd << ": ";
print(dp);
}
return 0;
}