Kruskal和map映射 解决POJ 2075

 

Description

You are the owner of SmallCableCo and have purchased the franchise rights for a small town. Unfortunately, you lack enough funds to start your business properly and are relying on parts you have found in an old warehouse you bought. Among your finds is a single spool of cable and a lot of connectors. You want to figure out whether you have enough cable to connect every house in town. You have a map of town with the distances for all the paths you may use to run your cable between the houses. You want to calculate the shortest length of cable you must have to connect all of the houses together.

Input

Only one town will be given in an input. 

• The first line gives the length of cable on the spool as a real number. 
• The second line contains the number of houses, N 
• The next N lines give the name of each house's owner. Each name consists of up to 20 characters {a–z,A–Z,0–9} and contains no whitespace or punctuation. 
• Next line: M, number of paths between houses 
• next M lines in the form

< house name A > < house name B > < distance > 
Where the two house names match two different names in the list above and the distance is a positive real number. There will not be two paths between the same pair of houses.

Output

The output will consist of a single line. If there is not enough cable to connect all of the houses in the town, output 
Not enough cable 
If there is enough cable, then output 
Need < X > miles of cable 
Print X to the nearest tenth of a mile (0.1).

Sample Input

100.0
4
Jones
Smiths
Howards
Wangs
5
Jones Smiths 2.0
Jones Howards 4.2
Jones Wangs 6.7
Howards Wangs 4.0
Smiths Wangs 10.0

Sample Output

Need 10.2 miles of cable

 

其实只要你会了Kruskal算法，那么这道上题目的关键就在

于映射了。为什么呢？因为要用Kruskal就必然要用并查

集，而并查集处理的是数字间的关系，不可以处理字符串的。

所以，在这道题目中要将不同的名称映射成不同的数字，以

此来用并查集判断是否存在回路。可以说，做好了映射，这

道题目相当的简单。那么怎样来映射呢？这里有一个十分有

效的工具，那就是map,可以说这个是对我在编程技巧栏目中

发表《映射在编程中的应用》的补充和完善。

#include "iostream"
#include "algorithm"
#include "cstring"
#include "string"
#include "map"
using namespace std;
struct line  //凡是用kruskal的结构体，其中的begin和end必是整形的，不然，无法使用并查集
{
    int begin;
    int end;
    double length;
};
map<string, int> s_to_int;
line num[1000];
int sum_line, father[1000], i, j;
int a, b;
double toatl, minlen;
int peo, group;
string name;
int find(int k)
{
    return father[k]==k?k:father[k]=find(father[k]);
}
int cmp(line first, line second)
{
    return first.length<second.length;
}
double kruskal()
{
    minlen = 0.0;
    for(i=0; i<peo; i++)
        father[i] = i;
    for(i=0; i<group; i++)
    {
        a = find(num[i].begin);
        b = find(num[i].end);
        if(a!=b)
        {
            father[a] = b;
            minlen += num[i].length;
        }
    }
    return minlen;
}
void init()
{
    cin>>toatl>>peo;
    for(i=0; i<peo; i++)
    {
        cin>>name;
        s_to_int[name] = i;
    }
    cin>>group;
    for(i=0; i<group; i++)
    {
        cin>>name;
        num[i].begin = s_to_int[name];
        cin>>name;
        num[i].end = s_to_int[name];
        cin>>num[i].length;
    }
    sort(num, num+group, cmp);
}
int main()
{
    init();
    if(kruskal()<=toatl)
        cout<<"Need "<<kruskal()<<" miles of cable"<<endl;
    else
        cout<<"Not enough cable"<<endl;
    return 0;
}