[POJ 1797] Heavy Transportation【Dijkstra】
Problem:
Time Limit: 3000MS | Memory Limit: 30000K |
Description
Hugo Heavy is happy. After the breakdown of the Cargolifter project he can now expand business. But he needs a clever man who tells him whether there really is a way from the place his customer has build his giant steel crane to the place where it is needed on which all streets can carry the weight.
Fortunately he already has a plan of the city with all streets and bridges and all the allowed weights.Unfortunately he has no idea how to find the the maximum weight capacity in order to tell his customer how heavy the crane may become. But you surely know.Problem
You are given the plan of the city, described by the streets (with weight limits) between the crossings, which are numbered from 1 to n. Your task is to find the maximum weight that can be transported from crossing 1 (Hugo's place) to crossing n (the customer's place). You may assume that there is at least one path. All streets can be travelled in both directions.
Input
Output
Sample Input
1 3 3 1 2 3 1 3 4 2 3 5
Sample Output
Scenario #1: 4
Source
Solution:
本题与[POJ 2253] Frogger【Dijkstra】一样水有异曲同工之妙,只是路径每段最小值变为最大值。
这次用邻接表实现,但在本题的条件下邻接表的时空优势不明显。
Code: O(Tn2) [1352K, 250MS]
#include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; int T, n, m; struct Edge{ int np, maxw; Edge *nxt; }; struct Graph{ Edge *V[1002], E[1000002]; int tope; inline void clear() {tope = 0, memset(V, 0, sizeof(V));} // Must set V[] to NULL inline void addedge(int u, int v, int w){ E[++tope].np = v, E[tope].maxw = w; E[tope].nxt = V[u], V[u] = &E[tope]; } } G; int maxload[1002]; bool vis[1002]; int main(){ scanf("%d", &T); for(register int cas = 1; cas <= T; cas++){ scanf("%d%d", &n, &m); G.clear(); for(register int i = 1; i <= m; i++){ int u, v, w; scanf("%d%d%d", &u, &v, &w); G.addedge(u, v, w), G.addedge(v, u, w); } memset(maxload, 0, sizeof(maxload)), maxload[1] = 0x3f3f3f3f; memset(vis, 0, sizeof(vis)); for(register int i = 1; i <= n; i++){ int maxl = 0; for(register int j = 1; j <= n; j++) if(!vis[j] && maxload[j] > maxload[maxl]) maxl = j; vis[maxl] = 1; // Find the max-loaded crossing to expand if(maxl == n) break; // Deliver goods to customer's door, well done! for(register Edge *ne = G.V[maxl]; ne; ne = ne->nxt) maxload[ne->np] = max(maxload[ne->np], min(maxload[maxl], ne->maxw)); // Update max load } printf("Scenario #%d:\n%d\n\n", cas, maxload[n]); } return 0; }
发表评论