Lift Hopping UVA - 10801 (建圖+Dijkstra)
阿新 • • 發佈:2018-12-26
題意:在一個大樓裡,有編號為0~99的100層樓裡,還有n座電梯,任務是從0樓到達第k樓,每個電梯都有一個執行速度,表示到達一個相鄰樓層需要的時間,由於每個電梯不一定每層都停留,有時需要從一個電梯換到另一個電梯中。換電梯時間總是一分鐘,前提兩座電梯能停留在換乘樓層,大樓裡只有你一個人,問最少需要的時間?
題解:對於電梯x來說,假如它在第i層停留,把x在i層的出口看作一個頂點記為x*100+i。對於x的下一個停靠站j來說,x*100+i和x*100+j可以連一條邊,權值就是這個電梯從i到j層所需要的時間,而對於同樣在i層停靠的每個電梯來說,x*100+i到y*100+i也連一條邊,權值為60(同層換乘)。建圖後,遍歷所有的x*100的點u和y*100+k的v,使用dijkstra演算法求出所有的u,v最短距離的最小值即為所求結果。
附上程式碼:
#include<bits/stdc++.h> #define _for(i,a,b) for( int i=(a); i<(b); ++i) #define _rep(i,a,b) for( int i=(a); i<=(b); ++i) using namespace std; const int MAXN = 5, MAXK = 100, MAXP = MAXN*MAXK; const int INF = 100000 + 10; struct Edge{ int from, to, dist; }; struct HeapNode{ int d, u; bool operator < (const HeapNode& rhs) const { return d > rhs.d; } }; template<int MAXSIZE> struct Dijkstra{ int n, m, d[MAXSIZE], p[MAXSIZE]; vector<Edge> edges; vector<int> G[MAXSIZE]; bool done[MAXSIZE]; void init(int n) { this->n = n; for(int i = 0; i < n; i++) G[i].clear(); edges.clear(); m = 0; } void addEdge(int from, int to, int dist) { edges.push_back((Edge){from, to, dist}); m = edges.size(); G[from].push_back(m - 1); } void dijkstra(int s) { priority_queue<HeapNode> Q; fill_n(d, n, INF); d[s] = 0; fill_n(done, n, false); Q.push((HeapNode){0, s}); while(!Q.empty()) { HeapNode x = Q.top(); Q.pop(); int u = x.u; if(done[u]) continue; done[u] = true; const vector<int>& adj = G[u]; for(int i = 0; i < adj.size(); i++) { Edge& e = edges[adj[i]]; int to = e.to; if(d[to] > d[u] + e.dist) { d[to] = d[u] + e.dist; p[to] = adj[i]; Q.push((HeapNode){d[to], to}); } } } } }; Dijkstra<MAXP+1> solver; int n, k, T[MAXN]; vector<int> Level[MAXK]; char buf[512]; int readint() { int x; scanf("%d", &x); return x; } int solve() { solver.init(n*MAXK+1); _for(i, 0, n) T[i] = readint(); fgets(buf,500,stdin); _for(e, 0, n){ fgets(buf,500,stdin); stringstream iss(buf); bool first = true; int l1, l; // 上一層,當前層 while(iss>>l) { if(first) first = false; else { int v = MAXK*e + l1, v2= MAXK*e + l; int dist = (l-l1)*T[e]; solver.addEdge(v,v2,dist); solver.addEdge(v2,v,dist); // printf("[%d,%d]-%d\n",v, v2, dist); } Level[l].push_back(e); l1 = l; } } _for(i, 0, MAXK) { vector<int>& li = Level[i]; _for(j, 0, li.size()){ _for(m, j+1, li.size()) { int e1 = li[j], e2 = li[m], v1 = e1*MAXK+i, v2 = e2*MAXK+i; solver.addEdge(v1,v2,60); solver.addEdge(v2,v1,60); // printf("[%d,%d]-%d\n",v1, v2, 60); } } } vector<int>& L0 = Level[0]; vector<int>& Lk = Level[k]; if(L0.empty() || Lk.empty()) return INF; int ans = INF; for(auto i : L0) { solver.dijkstra(i * 100); for(auto j : Lk) ans = min(ans, solver.d[j*100 + k]); } return ans; } int main(){ while(scanf("%d %d\n", &n, &k) == 2) { int ans = solve(); if(ans != INF) printf("%d\n", ans); else puts("IMPOSSIBLE"); } return 0; }