Lift Hopping UVA

对于可以通过同一个电梯到达的楼层与楼层之间建边，边的权值为电梯的运动所需要花费的时间的值，然后从第0层开始使用迪杰斯特拉算法得出最终的结果，这里需要注意几个问题：边是双向的，也就是对应电梯既可以从下向上运动也可以从上向下运动，同时对于目标楼层为第0层的情况要进行特判，对于可达到的情况我们需要减去一个60，因为开始进入电梯的时候是不要转移的时间的，但是为了在代码中方便进行统一的处理，我们都加上了60。其他的具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;class Edge{public:int from,to,time,next;Edge(int f = 0, int to = 0, int t = 0,int n=-1) :from(f), to(to), time(t),next(n){}Edge operator=(const Edge& a){from = a.from, to = a.to, time = a.time, next = a.next;return *this;}};Edge edge[50000];class Solve{public:int n, k;vector<int> t;vector<int> floor[110];int dist[110];bool visit[110];const int Inf = 0x3f3f3f3f;int next[110];int total;void addEdge(int from,int to,int perTime){edge[total] = Edge(from, to, abs(from - to)*perTime);edge[total].next = next[from];next[from] = total;total++;}void Init(){t.clear();total = 0;memset(next, -1, sizeof(next));for (int i = 0; i < 110; i++){floor[i].clear();}for (int i = 0; i < n; i++){int temp;cin >> temp;t.push_back(temp);}for (int i = 0; i < n; i++){while (true){int temp1;char temp2;cin >> temp1;temp2=getchar();floor[i].push_back(temp1);int amount = floor[i].size()-1;for (int k1 = 0; k1 < amount; k1++){addEdge(floor[i][k1], floor[i][amount], t[i]);addEdge(floor[i][amount], floor[i][k1], t[i]);}if (temp2 == '\n') break;}}}void Deal(){Init();memset(dist, Inf, sizeof(dist));memset(visit, 0, sizeof(visit));dist[0] = 0;for (int i = 0; i < 100; i++){int ind=-1, Min = Inf;for (int j = 0; j < 100; j++){if (!visit[j] && dist[j] < Min){Min = dist[j];ind = j;}}if (ind == -1) break;visit[ind] = true;for (int j = next[ind]; j != -1;j=edge[j].next){int from = edge[j].from;int to = edge[j].to;if (dist[from]<Inf && dist[to] > dist[from] + edge[j].time + 60){dist[to] = dist[from] + edge[j].time + 60;}}}if (k == 0) cout << "0\n";else if (dist[k] == Inf) cout << "IMPOSSIBLE\n";else cout << dist[k] - 60 << endl;}};int main(){Solve a;while (cin >> a.n >> a.k){a.Deal();}return 0;}