Description
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 5Sample Output
Scenario #1: 4注意题目是和求最短路是有点不同的。 /*审题，这个题是要求从1到n城市最大承载量，而最大承载量就是从城市1到城市n所有通路上的最大承载量，并不是求最短路， 不过，也可以用Dijkstra求解，只不过需要在求最短路上的Dijkstra模板上改改一些而已*/ #include<iostream> #include<cstring> #include<algorithm> #include<cstdio> using namespace std; const int maxn = 1010; int mp[maxn][maxn],dis[maxn],vis[maxn]; int edge,node; int num; void dijkstra(){ for(int i = 1; i <= node; i++){ dis[i] = mp[1][i]; } for(int i = 1; i < node; i++){ int mn = -1,u; for(int j = 1; j <= node; j++){ if(vis[j] == 0 && dis[j] > mn){ //dis[]是为了找到最大承载量。找到没有标记的点。 mn = dis[j]; u = j; } } vis[u] = 1; for(int j = 1; j <= node; j++){ if(vis[j] == 0 && dis[j] < min(dis[u],mp[u][j])){ //松弛，由于最大承载量是有限制，需要要选择最小的。 dis[j] = min(dis[u],mp[u][j]); } } } } int main(){ int T; num = 0; scanf("%d",&T); while(T--){ num++; scanf("%d%d",&node,&edge); for(int i = 1; i <= node; i++){ for(int j = 1; j <= node; j++){ mp[i][j] = 0; //求最大当然要初始化最小。 } } memset(vis,0,sizeof(vis)); int m,n,t; for(int i = 0; i < edge; i++){ scanf("%d%d%d",&m,&n,&t); mp[m][n] = mp[n][m] = t; //此题是无向图中求从1到node的最短路。无向图。 } dijkstra(); printf("Scenario #%d:\n%d\n\n",num,dis[node]); } return 0; } View Code
转载于:https://www.cnblogs.com/Weixu-Liu/p/10418170.html
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