Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem H - IP-TV

Background

A consortium of European Internet providers manages a large backbone network, with direct links (connections) between a large number of European cities. A link between a pair of cities is bidirectional. The transmission of a message in a link has an associated cost. As it is common in the Internet, it is possible to use a (unbounded) sequence of direct links to indirectly transfer data between any pair of cities.

For allowing the broadcast of TV programs using this backbone, it is necessary to continuously send data to all nodes in the network. For helping to minimize costs, it is necessary to select the network links that will be used for transmitting data. The set of selected links must be connected and include all nodes in the network.

For helping the consortium to manage its network, you have been asked to create a program that computes the minimum cost for transmitting data to all cities of the backbone.

Problem

Given a set of network links, compute the minimum transmission cost for reaching all nodes.

Input

Input consists of multiple test cases the first line of the input contains the number of test cases. There is a blank line before each dataset. The first line of each dataset contains a positive integer M, not greater than 2,000, with the number of cities that have network connections. The second line contains an integer N not greater than 50,000, with the number of existing links. Each of the following N lines contains the representation of a link. Each line contains two strings and one integer, separated by empty spaces, B E C, where B and E are the city names of the endpoints of the network link, with no more than 8 characters, and C is a positive integer, not greater than 30, representing the cost of transmitting in the link.

Output

The output consists of one single line that contains an integer with the minimum transmission cost for sending data to all cities. Print a blank line between datasets.

Sample Input

1

 

4

5

lisbon london 6

lisbon paris 5

london paris 1

london berlin 2

paris berlin 10

Sample Output

8

---

作法 : 最小生成樹
不會寫 HASH, 犧牲一下記憶體做個 Trie

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
typedef struct {
    int x, y, v;
}Path;
Path D[50001];
int cmp(const void *i, const void *j) {
    Path *a, *b;
    a = (Path *)i, b = (Path *)j;
    return a->v - b->v;
}
struct {
    int son[128];
    int code;
}Trie[60001];
int TrieSize;
int Recode(int *n, char *name) {
    static int i, j;
    for(i = 0, j = 0; name[i]; i++) {
        if(!Trie[j].son[name[i]])
            Trie[j].son[name[i]] = ++TrieSize;
        j = Trie[j].son[name[i]];
    }
    if(!Trie[j].code)
        Trie[j].code = (*n)++;
    return Trie[j].code;
}
int Parent[2001], Rank[2001];
int MakeSet(int N) {
    static int i;
    for(i = 1; i <= N; i++)
        Parent[i] = i, Rank[i] = 1;
}
int FindP(int x) {
    if(x != Parent[x])
        Parent[x] = FindP(Parent[x]);
    return Parent[x];
}
int Link(int x, int y) {
    if(Rank[x] < Rank[y])
        Parent[x] = y, Rank[x] += Rank[y];
    else
        Parent[y] = x, Rank[y] += Rank[x];
}
int Union(int x, int y) {
    x = FindP(x), y = FindP(y);
    if(x != y) {
        Link(x, y);
        return 1;
    }
    return 0;
}
int main() {
    int T, M, N, C, size, i, x, y;
    char B[31], E[31];
    scanf("%d", &T);
    while(T--) {
        size = 1, TrieSize = 0;
        scanf("%d %d", &M, &N);
        memset(Trie, 0, sizeof(Trie));
        for(i = 0; i < N; i++) {
            scanf("%s %s %d", B, E, &C);
            x = Recode(&size, B);
            y = Recode(&size, E);
            D[i].x = x, D[i].y = y, D[i].v = C;
        }
        qsort(D, N, sizeof(Path), cmp);
        MakeSet(M);
        int Ans = 0;
        for(i = 0; i < N; i++)
            Ans += D[i].v * Union(D[i].x, D[i].y);
        printf("%d\n", Ans);
        if(T)    puts("");
    }
    return 0;
}