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

編輯：關於C++

Tangled in Cables Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6039 Accepted: 2386

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

Source

Mid-Atlantic 2004

題目大意：給一個總線長，和n個人，m組關系，表示兩個人之間的距離，現在要求所有人之間都連通，問最少需要多長的線，若超過總長，則輸出Not enough cable

題目分析：裸最小生成樹問題，直接map一下字符串，再跑一下Kruskal

#include 
#include 
#include
#include 
#include 
using namespace std;
int const MAX = 1e4;

int fa[MAX];
int re[MAX];
int n, m;
map 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
, ans);
    else
        printf(Not enough cable
);
}