Bzoj4753/洛谷P4432 [JSOI2016]最佳团体（0/1分数规划+树形DP）

题面

Bzoj

洛谷

题解

这种求比值最大就是$0/1$分数规划的一般模型。

这里用二分法来求解最大比值，接着考虑如何$check$，这里很明显可以想到用树形背包$check$，但是时间复杂度要优化成$O(n^2)$的，可以参考之前写的这篇博客

#include <cstdio>
#include <algorithm>
using std::min;
using std::max;

const int N = 3e3 + 10, inf = 1e9 + 7;
const double eps = 1e-5;
int n, K, s[N], p[N], son[N][N], dfn[N], time, nx[N];
int from[N], to[N], nxt[N], cnt;//Edges
double f[N][N], d[N];

inline void addEdge (int u, int v) {
    to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}

inline void upt(double &a, double b) {
    if (a < b) a = b;
}

void dfs (int u) {
    dfn[u] = time++;
    for (int i = from[u]; i; i = nxt[i]) dfs(to[i]);
    nx[dfn[u]] = time;
}

inline bool check (double k) {
    for (int i = 1; i <= n; ++i)
        d[dfn[i]] = p[i] - k * s[i];
    for (int i = 1; i <= n + 1; ++i)
        for (int j = 0; j <= K; ++j)
            f[i][j] = -inf;
    for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= min(i, K); ++j) {
            upt(f[i + 1][j + 1], f[i][j] + d[i]);
            upt(f[nx[i]][j], f[i][j]);
        }
    return f[n + 1][K] >= eps;
}

int main () {
    scanf("%d%d", &K, &n); ++K;
    for (int i = 1, fa; i <= n; ++i)  {
        scanf("%d%d%d", s + i, p + i, &fa);
        addEdge(fa, i);
    }
    dfs(0);
    double l = 0, r = 10000, ans;
    while (r - l >= eps) {
        double mid = (l + r) * 0.5;
        if (check(mid)) ans = mid, l = mid + eps;
        else r = mid - eps;
    }
    printf ("%.3lf\n", ans);
    return 0;
}