POJ 2075 Tangled in Cables (kruskal算法 MST + map)

• 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

#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#include <iostream>
using namespace std;
int const MAX = 1e4;

int fa[MAX];
int re[MAX];
int n, m;
map<string, int> mp;

struct Edge
{
    int u, v;
    double w;
}e[MAX];

bool cmp(Edge a, Edge b)
{
    return a.w < b.w;
}

void UF_set()
{
    for(int i = 0; i < MAX; i++)
        fa[i] = i;
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void Union(int a, int b)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(r1 != r2)
        fa[r2] =r1;
}

double Kruskal()
{
    UF_set();
    int num = 0;
    double res = 0;
    for(int i = 0; i < m; i++)
    {
        int u = e[i].u;
        int v = e[i].v;
        if(Find(u) != Find(v))
        {
            Union(u, v);
            res += e[i].w;
            num ++;
        }
        if(num >= n - 1)
            break;
    }
    return res;
}

int main()
{
    string s, s1, s2;
    double val, sum, ans = 0;
    scanf("%lf", &sum);
    int cnt = 1;
    mp.clear();
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        cin >> s;
        if(!mp[s])
            mp[s] = cnt ++;
    }
    scanf("%d", &m);
    for(int i = 0; i < m; i++)
    {
        cin >> s1 >> s2 >> val;
        e[i].u = mp[s1];
        e[i].v = mp[s2];        
        e[i].w = val;
    }
    sort(e, e + m, cmp);
    ans = Kruskal();
    if(ans < sum)
        printf("Need %.1f miles of cable\n", ans);
    else
        printf("Not enough cable\n");
}