原題鏈接
Background 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 The first line contains the number of scenarios (city plans). For each city the number n of street crossings (1 <= n <= 1000) and number m of streets are given on the first line. The following m lines contain triples of integers specifying start and end crossing of the street and the maximum allowed weight, which is positive and not larger than 1000000. There will be at most one street between each pair of crossings.
Output The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the maximum allowed weight that Hugo can transport to the customer. Terminate the output for the scenario with a blank line.
Sample Input 1 3 3 1 2 3 1 3 4 2 3 5
Sample Output Scenario #1: 4
思路1：dijkstra算法 d[i]不再是從起點到i點的最短距離，而是從起點到i點過程中的最小承重限制。 不斷從未訪問的點中挑選出d[v]最大的點v，再更新其他點的d[i]值，即d[i]=max（d[i]，min（d[v]，weight[v][i]））
AC代碼：
#include <iostream>#include <cstdio> #include <cstring> #include <cstdlib>#include <cmath>#include <algorithm> using namespace std; int t, n, m, weight[1005][1005], vis[1005], d[1005];int dijkstra(){ int i, j; fill(vis + 1, vis + 1 + n, 0); for(i = 1; i <= n; i++){ d[i] = weight[1][i]; } for(i = 1; i <= n; i++){ int max_weight = -1; int v; for(j = 1; j <= n; j++){ if(!vis[j] && d[j] > max_weight){ max_weight = d[j]; v = j; } } vis[v] = 1; for(j = 1; j <= n; j++){ if(!vis[j] && d[j] < min(d[v], weight[v][j])){ d[j] = min(d[v], weight[v][j]); } } } return d[n]; } int main(){ int i, j, k; scanf("%d", &t); for(k = 1; k <= t; k++){ scanf("%d %d", &n, &m); for(i = 1; i <= n; i++) fill(weight[i] + 1, weight[i] + 1 + n, 0); for(i = 1; i <= m; i++){ int a, b, c; scanf("%d %d %d", &a, &b, &c); weight[a][b] = weight[b][a]=c; } printf("Scenario #%d:/n%d/n/n", k, dijkstra()); } return 0; }思路2：kruskal算法 從權值最大的邊開始，依次判斷它是否能加入并查集，直到起點和終點連通為止，此時加入的最后一條邊的權值是所求答案。
AC代碼：
#include<cstdio>#include<algorithm>using namespace std;struct Edge{ int from, to, cost;}es[1000005]; int N,M;int par[1005];int rank[1005];//并查集 void init(int n){ for(int i = 1; i <= n; ++i){ par[i] = i; rank[i] = 1; }}int find(int x){ if(x == par[x]) return x; else return par[x] = find(par[x]);}void unite(int x, int y){ x = find(x); y = find(y); if(x == y) return; if(rank[x] < rank[y]) par[x] = y; else{ par[y] = x; if(rank[x] == rank[y]) rank[x]++; }}bool same(int x, int y){ return find(x) == find(y); }bool EdgeCmp(Edge &e1, Edge &e2){ return e1.cost > e2.cost; }int Kruskal(){ sort(es, es + M, EdgeCmp); init(N); for(int i = 0; i < M; ++i){ Edge &e = es[i]; if(!same(e.from, e.to)){ unite(e.from, e.to); if(same(1, N)) return e.cost; } } return -1;}int main(){ int T; scanf("%d",&T); for(int t = 1; t <= T; ++t){ scanf("%d %d", &N, &M); for(int m = 0; m < M; ++m) scanf("%d %d %d", &es[m].from, &es[m].to, &es[m].cost); printf("Scenario #%d:/n%d/n/n", t, Kruskal()); //注意題目要求留空白行，所以是/n/n } return 0;}新聞熱點
疑難解答
