POJ 2718 分数规划

链接：

http://poj.org/problem?id=2728

题意：

给你n个点，每两个点之间有value和cost，找一个最小生成树，使得∑cost/∑value最小

题解：

典型的分数规划题，二分+最小生成树就可以了

代码：

struct point { double x, y, z; }p[MAXN];
double a[MAXN][MAXN];
double b[MAXN][MAXN];
double cost[MAXN][MAXN];
double d[MAXN];
bool used[MAXN];
int n;

double prim() {
    rep(i, 0, n) d[i] = 1e20;
    memset(used, 0, sizeof(used));
    d[0] = 0;
    double res = 0;

    while (1) {
        int v = -1;
        rep(u, 0, n) if (!used[u] && (v == -1 || d[u] < d[v])) v = u;
        if (v == -1) break; used[v] = 1; res += d[v];
        rep(u, 0, n) d[u] = min(d[u], cost[v][u]);
    }
    return res;
}

double sqr(double x) {
    return x*x;
}

bool check(double x) {
    rep(i, 0, n) rep(j, 0, n)
        cost[i][j] = -x * a[i][j] + b[i][j];
    return prim() >= 0;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(0);
    while (cin >> n, n) {
        rep(i, 0, n) cin >> p[i].x >> p[i].y >> p[i].z;
        rep(i, 0, n) rep(j, 0, n) {
            a[i][j] = sqrt(sqr(p[i].x - p[j].x) + sqr(p[i].y - p[j].y));
            b[i][j] = fabs(p[i].z - p[j].z);
        }
        double l = 0, r = 100;
        rep(i, 0, 20) {
            double m = (l + r) / 2;
            if (check(m)) l = m;
            else r = m;
        }
        printf("%.3f
", l);
    }
    return 0;
}