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

Singapore will host a Formula One race in 2008. The race will be held on a 5.067km long street circuit, consisting of 14 left hand turns and 10 right hand turns. In the run up to the F1 race, the number of illegal night street racing activities have been on the rise. Such races consists of several rounds around a designated street circuit.

The authorities would like to deploy a new vehicle monitoring system in order to catch these illegal Saint Andrew's Road, part of the Formula One circuit racers. The system consists of a (Kenny Pek, Piccom) number of cameras mounted along various roads. For the system to be effective, there should be at least one camera along each of the possible circuits.

The Singapore road system can be represented as a series of junctions and connecting bidirectional roads (see Figure 5). A possible racing circuit consists of a start junction followed by a path consisting of three or more roads that eventually leads back to the start junction. Each road in a racing circuit can be traversed only in one direction, and only once.

Your task is to write a program that computes the optimal placement of the vehicle-monitoring cameras. You will be provided with a description of a connected road network to be monitored in terms of the roads and junctions. The junctions are identified by the bigger numbers in Figure 5. A camera can be deployed on the roads (and not the junctions).

The cost of deploying a camera depends on the road on which it is placed. The smaller numbers by the roads in Figure 5 indicate the cost of deploying a camera on that road. Your job is to select a set of roads that minimizes the total cost of deployment while ensuring that there is at least one camera along every possible racing circuit (i.e. loop in the road network).

Input 

The input consists of a line containing the number c of datasets, followed by c datasets, followed by a line containing the number `0'.

The first line of each dataset contains two positive integers, n and m , separated by a blank, which represent the number of junctions and number of roads, respectively. You may assume that 0 < n < 10000 and 0 < m < 100000 . For simplicity, we label each of the n junctions from 1 to n . The following m lines of each dataset each describes one road. Each line consists of three positive integers which are the labels of two different junctions and the cost of deploying a camera on this road. The cost of deploying a camera is between 1 and 1000.

Output 

The output consists of one line for each dataset. The c -th line contains one single number, representing the minimal cost of setting up the vehicle monitoring system such that there is at least one camera along every possible circuit.

Note: This data set depicts the situation shown in Figure 5. The two cameras show where cameras might be placed in order to monitor each circuit at minimal cost. Since each of the cameras have a cost of 3, the total minimal cost is 6.

Sample Input 

1 
6 7 
1 2 5
2 3 3
1 4 5
4 5 4
5 6 4
6 3 3
5 2 3
0

Sample Output 

6

---

題目的意思為賽道是一個環道, 但我們不知道賽道是怎麼分布的, 因此希望每個環道至少有一台監視器,
希望總成本最低, 那如果把環道剔除的話, 所剩的是一棵樹, 成本最低, 也就是希望產生的樹成本最高

#include <stdio.h>
struct edge {
    int x, y, v;
};
edge D[100000], E[100000];
void sort(int n) {
    int w[1001] = {}, i;
    for(i = 0; i < n; i++)
        w[D[i].v]++;
    for(i = 999; i >= 0; i--)
        w[i] += w[i+1];
    for(i = 0; i < n; i++) {
        E[--w[D[i].v]] = D[i];
    }
}
int p[10001], r[10001];
void init(int n) {
    int i;
    for(i = 0; i <= n; i++)
        p[i] = i, r[i] = 1;
}
int find(int x) {
    return p[x] == x ? x : (p[x]=find(p[x]));
}
int joint(int x, int y) {
    x = find(x), y = find(y);
    if(x != y) {
        if(r[x] > r[y])
            r[x] += r[y], p[y] = x;
        else
            r[y] += r[x], p[x] = y;
        return 1;
    }
    return 0;
}
int main() {
    int t, n, m, i;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        int sum = 0;
        for(i = 0; i < m; i++) {
            scanf("%d %d %d", &D[i].x, &D[i].y, &D[i].v);
            sum += D[i].v;
        }
        sort(m);
        init(n);
        for(i = 0; i < m; i++) {
            if(joint(E[i].x, E[i].y))
                sum -= E[i].v;
        }
        printf("%d\n", sum);
    }
    scanf("%d", &t);
    return 0;
}