POJ

Heavy Transportation

Time Limit: 3000MS
Memory Limit: 30000KTotal Submissions: 39515
Accepted: 10387

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

13 31 2 31 3 42 3 5

Sample Output

Scenario #1:4

Source

TUD Programming Contest 2004, Darmstadt, Germany

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题意：

求车的最大载重。

思路：

一辆车从1到n，求到达n的子路段中的最大载重 （不超出路的载重），即最长路的最小权值

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define INF 0x3f3f3f3fint n,m;int dis[1005][1005],dist[1005];int vis[1005];void dijkstra(){int ans=INF;memset(vis,0,sizeof(vis));for(int i=1;i<=n;i++)dist[i]=dis[1][i];vis[1]=1;m=n-1;while(m--){int maxn=0;int p=0;for(int i=1;i<=n;i++){if(!vis[i] && dist[i]>maxn){maxn=dist[i];p=i;}}if(p==0)break;vis[p]=1;for(int i=1;i<=n;i++){//求找到的p点相连的两段路中的最小值，因为当车的载重大于路的载重时 // 不能通行，所以要找此路中的最小值 if(!vis[i] && dist[i] < min(dis[p][i],dist[p])){dist[i]=min(dis[p][i],dist[p]);}}}printf("%d\n\n",dist[n]);}int main(){int t;scanf("%d",&t);for(int k=1;k<=t;k++){scanf("%d%d",&n,&m);memset(dis,0,sizeof(dis));for(int i=0;i<m;i++){int a,b,c;scanf("%d%d%d",&a,&b,&c);dis[a][b]=dis[b][a]=max(dis[a][b],c);}printf("Scenario #%d:\n",k);dijkstra();}return 0;}