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

A

Airports

 

 

The government of a certain developing nation wants to improve transportation in one of its most inaccessible areas, in an attempt to attract investment. The region consists of several important locations that must have access to an airport.

 

Of course, one option is to build an airport in each of these places, but it may turn out to be cheaper to build fewer airports and have roads link them to all of the other locations. Since these are long distance roads connecting major locations in the country (e.g. cities, large villages, industrial areas), all roads are two-way. Also, there may be more than one direct road possible between two areas. This is because there may be several ways to link two areas (e.g. one road tunnels through a mountain while the other goes around it etc.) with possibly differing costs.

 

A location is considered to have access to an airport either if it contains an airport or if it is possible to travel by road to another location from there that has an airport.

 

You are given the cost of building an airport and a list of possible roads between pairs of locations and their corresponding costs. The government now needs your help to decide on the cheapest way of ensuring that every location has access to an airport. The aim is to make airport access as easy as possible, so if there are several ways of getting the minimal cost, choose the one that has the most airports.

 

Note: The input file is large; make sure your I/O code is fast.

 

 

Input

 

The first line of input contains the integer T(T<25), the number of test cases. The rest of the input consists of T cases.

 

Each case starts with three integers N, M and A (0<N<=10,000, 0<=M<=100,000, 0<A<=10,000) separated by white space. N is the number of locations, M is the number of possible roads that can be built, and A is the cost of building an airport.

 

The following M lines each contain three integers X, Y and C (1<=X, Y<=N, 0<C<=10,000), separated by white space. X and Y are two locations, and C is the cost of building a road between X and Y.

 

Output

 

Your program should output exactly T lines, one for each case. Each line should be of the form "Case #X: Y Z", where X is the case number Y is the minimum cost of making roads and airports so that all locations have access to at least one airport, and Z is the number of airports to be built. As mentioned earlier, if there are several answers with minimal cost, choose the one that maximizes the number of airports.

 

 

Sample Input	Sample Output
2 4 4 100 1 2 10 4 3 12 4 1 41 2 3 23 5 3 1000 1 2 20 4 5 40 3 2 30	Case #1: 145 1 Case #2: 2090 2

---

作法 : MST (最小生成樹),  Kruskal算法
使用"并查集" Disjoint Sets

#include<stdio.h>
#include<stdlib.h>
typedef struct {
    int x, y, v;
}Edge;
int cmp(const void *i, const void *j) {
    Edge *a, *b;
    a = (Edge *)i, b = (Edge *)j;
    return a->v - b->v;
}
int P[10001], R[10001];
void MakeSet(int n) {
    static int i;
    for(i = 0; i <= n; i++)
        P[i] = i, R[i] = 1;
}
int FindP(int x) {
    if(P[x] != x)
        P[x] = FindP(P[x]);
    return P[x];
}
void Link(int x, int y) {
    if(R[x] < R[y])
        P[x] = y, R[y] += R[x];
    else
        P[y] = x, R[x] += R[y];
}
int Union(int x, int y, int *val) {
    x = FindP(x), y = FindP(y);
    if(x != y) {
        Link(x, y);
        (*val) = (*val) - 1;
        return 1;
    }
    return 0;
}
inline int readchar() {
    const int N = 1048576;
    static char buf[N];
    static char *p = buf, *end = buf;
    if(p == end) {
        if((end = buf + fread(buf, 1, N, stdin)) == buf) return EOF;
        p = buf;
    }
    return *p++;
}
inline int ReadInt(int *x) {
    static char c, neg;
    while((c = readchar()) < '-')    {if(c == EOF) return EOF;}
    neg = (c == '-') ? -1 : 1;
    *x = (neg == 1) ? c-'0' : 0;
    while((c = readchar()) >= '0')
        *x = (*x << 3) + (*x << 1) + c-'0';
    *x *= neg;
    return 1;
}
int main() {
    int t, N, M, A, i, C = 0, tM;
    Edge D[100000];
    ReadInt(&t);
    while(t--) {
        ReadInt(&N), ReadInt(&M), ReadInt(&A);
        for(i = 0, tM = 0; i < M; i++) {
            ReadInt(&D[tM].x), ReadInt(&D[tM].y), ReadInt(&D[tM].v);
            if(D[tM].v < A)    tM++;
        }
        qsort(D, tM, sizeof(Edge), cmp);
        /*Disjoint Sets*/
        MakeSet(N);
        int val = N, cost = 0;
        for(i = 0; i < tM; i++)
            cost += D[i].v*Union(D[i].x, D[i].y, &val);
        printf("Case #%d: %d %d\n", ++C,cost+val*A, val);
    }
    return 0;
}