POJ1797,Heavy Transportation,貌似是最大流什么的,有点特殊,所以拿Kruskal练练手

Heavy Transportation

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 5

Sample Output

Scenario #1:
4

分析:

题意是从图的一个点到另一个点,通过有限条路径,每条边有一个承重上限,求最优情况.

开始题意理解有些偏差,直接改一下Kruskal的比较函数cmp直接求最大生成树了,也没测试直接就交了...坑了几个WA...

只给出路口的范围,边的范围可以大概推出来

其实只要判断1点跟n点是否已经在同一个并查集中就可...不必求出整棵生成树...也就是说可以有些路口不必经过...还有每个case结束之后要多输出一个空行

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#define MAX 1000
#define INF 0x7fffffff
using namespace std;
int m,n;
int p[MAX+10],r[105010];
typedef struct edge
{
    int u,v,len;
}edge;
edge eg[105010];
bool cmp(int i,int j)
{
    return eg[i].len>eg[j].len;
}
int find(int x)
{
    return p[x]==x?x:p[x]=find(p[x]);
}
int Kruskal()
{
    int MIN=INF;
    for(int i=1;i<=n;i++) p[i]=i;
    for(int i=1;i<=m;i++)
    {
        scanf("%d %d %d",&eg[i].u,&eg[i].v,&eg[i].len);
        r[i]=i;
    }
    sort(r+1,r+m+1,cmp);
    for(int i=1;i<=m;i++)
    {
        if(find(1)!=find(n))
        {
            int tmp=r[i];
            MIN=eg[tmp].len;
            int pu=find(eg[tmp].u),pv=find(eg[tmp].v);
            if(pu!=pv) p[pv]=pu;
        }
        else break;
    }
    return MIN;
}
int main()
{
    //freopen("input.txt","r",stdin);
    int t,count;
    scanf("%d",&t);
    for(count=1;count<=t;count++)
    {
        scanf("%d %d",&n,&m);
        int ans=Kruskal();
        printf("Scenario #%d:\n%d\n\n",count,ans);
    }
    return 0;
}