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 E
Connect the Campus
Input: standard input
Output: standard output
Time Limit: 2 seconds

Many new buildings are under construction on the campus of the University of Waterloo. The university has hired bricklayers, electricians, plumbers, and a computer programmer. A computer programmer? Yes, you have been hired to ensure that each building is connected to every other building (directly or indirectly) through the campus network of communication cables.

We will treat each building as a point specified by an x-coordinate and a y-coordinate. Each communication cable connects exactly two buildings, following a straight line between the buildings. Information travels along a cable in both directions. Cables can freely cross each other, but they are only connected together at their endpoints (at buildings).

You have been given a campus map which shows the locations of all buildings and existing communication cables. You must not alter the existing cables. Determine where to install new communication cables so that all buildings are connected. Of course, the university wants you to minimize the amount of new cable that you use.

Fig: University of Waterloo Campus

Input

The input file describes several test case.  The description of each test case is given below:

The first line of each test case contains the number of buildings N (1<=N<=750). The buildings are labeled from 1 to N. The next N lines give the x and y coordinates of the buildings. These coordinates are integers with absolute values at most 10000. No two buildings occupy the same point. After that there is a line containing the number of existing cables M (0 <= M <= 1000) followed by M lines describing the existing cables. Each cable is represented by two integers: the building numbers which are directly connected by the cable. There is at most one cable directly connecting each pair of buildings.

Output

For each set of input, output in a single line the total length of the new cables that you plan to use, rounded to two decimal places.

Sample Input

4
103 104
104 100
104 103
100 100
1
4 2

4
103 104

104 100

104 103

100 100

1

4 2

Sample Output
4.41
4.41

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
struct edge {
    int x, y, v;
};
int p[751], r[751];
void init(int n) {
    for(int i= 0; i <= n; i++)
        p[i] = i, r[i] = 1;
}
int find(int x) {
    return x == p[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 cmp(const void *i, const void *j) {
    static edge *a, *b;
    a = (edge *)i, b = (edge *)j;
    return a->v - b->v;
}
edge D[500000];
int main() {
    int n, m, i, j;
    int x[750], y[750];
    while(scanf("%d", &n) == 1) {
        init(n);
        for(i = 0; i < n; i++) {
            scanf("%d %d", x+i, y+i);
        }
        scanf("%d", &m);
        int connect = 0;
        while(m--) {
            scanf("%d %d", &i, &j);
            i--, j--;
            connect += joint(i, j);
        }
        m = 0;
        for(i = 0; i < n; i++) {
            for(j = i+1; j < n; j++) {
                if(find(i) != find(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(edge), cmp);
        double ans = 0;
        for(i = 0; i < m; i++) {
            if(joint(D[i].x, D[i].y)) {
                ans += sqrt(D[i].v);
                connect++;
                if(connect == n-1)
                    break;
            }
        }
        printf("%.2lf\n", ans);
    }
    return 0;
}