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 C: Arctic Network

The Department of National Defence (DND) wishes to connect several northern outposts by a wireless network. Two different communication technologies are to be used in establishing the network: every outpost will have a radio transceiver and some outposts will in addition have a satellite channel.

Any two outposts with a satellite channel can communicate via the satellite, regardless of their location. Otherwise, two outposts can communicate by radio only if the distance between them does not exceed D, which depends of the power of the transceivers. Higher power yields higher D but costs more. Due to purchasing and maintenance considerations, the transceivers at the outposts must be identical; that is, the value of D is the same for every pair of outposts.

Your job is to determine the minimum D required for the transceivers. There must be at least one communication path (direct or indirect) between every pair of outposts.

The first line of input contains N, the number of test cases. The first line of each test case contains 1 <= S <= 100, the number of satellite channels, and S < P <= 500, the number of outposts. P lines follow, giving the (x,y) coordinates of each outpost in km (coordinates are integers between 0 and 10,000). For each case, output should consist of a single line giving the minimum D required to connect the network. Output should be specified to 2 decimal points.

Sample Input

1
2 4
0 100
0 300
0 600
150 750

Sample Output

212.13

---

求最小生成樹的最大邊, 但又衛星可以直接協助連接, 因此可以少算 S 個邊

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct E {
    int x, y;
    int v;
};
E D[250000];
int cmp(const void *i, const void *j) {
    E *a, *b;
    a = (E *)i, b = (E *)j;
    return a->v - b->v;
}
int p[501], r[501];
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])
            p[y] = x, r[x] += r[y];
        else
            p[x] = y, r[y] += r[x];
        return 1;
    }
    return 0;
}
int main() {
    int t, S, P;
    int x[501], y[501], i, j;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &S, &P);
        int m = 0;
        for(i = 0; i < P; i++) {
            scanf("%d %d", &x[i], &y[i]);
            for(j = i-1; j >= 0; j--) {
                D[m].x = i, D[m].y = j;
                D[m].v = (x[i]-x[j])*(x[i]-x[j]) + (y[i]-y[j])*(y[i]-y[j]);
                m++;
            }
        }
        qsort(D, m, sizeof(E), cmp);
        init(P);
        int cnt = 0;
        for(i = 0; i < m; i++) {
            cnt += joint(D[i].x, D[i].y);
            if(cnt == P-S) {
                printf("%.2lf\n", sqrt(D[i].v));
                break;
            }
        }
    }
    return 0;
}