CUGB图论专场:C - Tangled in Cables(最小生成树kruscal)

C - Tangled in Cables
Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u
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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

纯粹的最小生成树,只不过结点变成了字符串而已,可以用STL中的map转化成数字型的结点……

#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <list>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#define PI acos(-1.0)
#define mem(a,b) memset(a,b,sizeof(a))
#define sca(a) scanf("%d",&a)
#define M 10010
#define INF 10000000
using namespace std;
typedef long long ll;
map<string,int>q;
int i,n,m,f[M];
struct Edge
{
    int u,v;
    double len;
} e[M];
double cmp(Edge a, Edge b)
{
    return a.len<b.len;
}
int find(int k)
{
    return f[k]==k?k:f[k]=find(f[k]);
}
double Kruskal()
{
    int j=0,a,b;
    double sum=0;
    sort(e,e+m,cmp);
    for(i=1;i<=n;i++)//把各个顶点归到不同的集合
        f[i]=i;
    for(i=0;i<m;i++)
    {
        a = find(e[i].u);
        b = find(e[i].v);
        if (a != b)//若不在同一集合,则MST中包含该边
        {
            f[b] = a;
            j++;
            sum += e[i].len;
            if(j == n - 1) break;//n个顶点的生成树有且只有n-1条边,达到即可退出
        }
    }
    return sum;
}
int main()
{
    double l,ll,abc;
    string a,b;
    cin>>l>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a;
        q[a]=i;
    }
    cin>>m;
    for(i=0;i<m;i++)
    {
        cin>>a>>b>>ll;
        e[i].u=q[a];e[i].v=q[b];e[i].len=ll;
    }
    abc=Kruskal();
    if(abc<=l) printf("Need %.1f miles of cable\n",abc);
    else printf("Not enough cable\n");
    return 0;
}


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