POJ 1797 Heavy Transportation

Heavy Transportation

Time Limit: 3000MS Memory Limit: 30000KTotal Submissions: 40918 Accepted: 10763

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

题意：有n个城市，m条道路，在每条道路上有一个承载量，现在要求从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。

解题思路：没必要把所有的点全部连起来，跑一遍最大生成树（也没必要跑完），每次更新并查集时判断1和n是否在同一个并查集内，如果在直接输出就行，因为此时选到的边一定是这个生成树内最小的边。附上代码

#include<stdio.h>#include<iostream>using namespace std;#include<algorithm>#include<string.h>#define maxn 1005#define inf 0x3f3f3f3fint n,m;struct road{    int from,to,val;} roads[maxn*maxn/2];int p[maxn];int cmp(road a,road b){    return a.val>b.val;}int find(int x){    return p[x]==x?x:find(p[x]);}int kruskal(){    for(int i=1; i<=n; i++)        p[i]=i;    for(int i=0; i<m; i++)    {        int x=find(roads[i].from);        int y=find(roads[i].to);        if(x!=y)            p[x]=y;        if(find(1)==find(n))     /*这里是最重要的*/            return roads[i].val;    }}int main(){    //freopen("in.txt","r",stdin);    int t;    cin>>t;    for(int k=1;k<=t;k++)    {        printf("Scenario #%d:\n",k);        scanf("%d%d",&n,&m);        for(int i=0; i<m; i++)            scanf("%d%d%d",&roads[i].from,&roads[i].to,&roads[i].val);        sort(roads,roads+m,cmp);        printf("%d\n\n",kruskal());    }    return 0;}