【最小生成树】Building a Space Station POJ

Building a Space Station POJ - 2031

题意：

给定n个球形空间的坐标(x,y,z)以及半径。以下三种情况均可视为空间之间连通：（1）两球之间有重合部分或一个被另一个完全包含在内；（2）两球之间有走廊直接连接；（3）两球之间通过走廊间接连接。走廊建在两个球形空间的表面上。问连通所有空间的走廊的最小长度。

思路：

直接套最小生成树的板子。注意建双向边。

maxn又又又又开小了，只开了500+100这么大，又用这个数来开存边的数组……还好G++会提示RE，这才反应过来

const int maxn = 50000 + 100;

int fa[maxn];
int tmp[maxn];
double x[maxn], y[maxn], z[maxn], r[maxn], w[maxn];
int u[maxn], v[maxn];
int n, m;

int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool cmp(int i, int j) {
    return w[i] < w[j];
    //if (w[i] - w[j] < -eps) return true;
    //else return false;
};

double count(double x1, double y1, double z1, double x2, double y2, double z2,double r1,double r2) {
    double dx = (x1 - x2) * (x1 - x2);
    double dy = (y1 - y2) * (y1 - y2);
    double dz = (z1 - z2) * (z1 - z2);
    double dr = r1 + r2;
    if (sqrt(dx + dy + dz) <= dr) return 0;
    else return sqrt(dx + dy + dz) - dr;
}

double solve() {
    double ans = 0;
    for (int i = 0; i < maxn; i++) fa[i] = i;
    for (int i = 0; i < m; i++) tmp[i] = i;
    sort(tmp, tmp + m, cmp);
    for (int i = 0; i < m; i++) {
        int e = tmp[i];
        int from = find(u[e]);
        int to = find(v[e]);
        if (from != to) {
            ans += w[e];
            fa[from] = to;
        }
    }
    return ans;
}

int main()
{
    //ios::sync_with_stdio(false);
    while (cin >> n && n) {
        m = 0;
        for (int i = 1; i <= n; i++) {
            cin >> x[i] >> y[i] >> z[i] >> r[i];
            for (int j = i; j >= 1; j--) {
                double cost = count(x[i],y[i],z[i],x[j],y[j],z[j],r[i],r[j]);
                w[m] = cost; u[m] = i; v[m] = j; m++;
                w[m] = cost; u[m] = j; v[m] = i; m++;
            }
        }
        printf("%.3f
", solve());
    }
    return 0;
}