[JSOI 2016] 最佳团体

Posted 2021-01-10 evenbao

[题目链接]

        https://www.lydsy.com/JudgeOnline/problem.php?id=4753

[算法]

         很明显的分数规划

         可以用树形动态规划（树形背包）检验答案

         时间复杂度 ： O(N^3logN)

[代码]

      

#include<bits/stdc++.h>
using namespace std;
#define MAXN 2510
const double eps = 1e-4;
const double inf = 1e9;

int n , tot , k;
int head[MAXN],a[MAXN],b[MAXN],size[MAXN],father[MAXN];
double f[MAXN][MAXN];
double value[MAXN],tmp[MAXN];

struct edge
{
        int to , nxt;
} e[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline void dp(int u)
{
        size[u] = 1;
        f[u][0] = 0;
        f[u][1] = value[u];
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                dp(v);
                for (int j = 1; j <= size[u] + size[v]; j++) tmp[j] = -inf;
                for (int j = 1; j <= size[u]; j++)
                {
                        for (int k = 0; k <= size[v]; k++)
                        {
                                chkmax(tmp[j + k],f[u][j] + f[v][k]);
                        }
                }
                for (int j = 1; j <= size[u] + size[v]; j++) f[u][j] = tmp[j];
                size[u] += size[v];
        }
}
inline void addedge(int u,int v)
{
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] = tot;
}
inline bool check(double mid)
{
        for (int i = 1; i <= n; i++) value[i] = (double)1.0 * b[i] - (double)1.0 * mid * a[i];
        for (int i = 0; i <= n; i++) 
        {
                for (int j = 0; j <= n + 1; j++)
                {
                        f[i][j] = -inf;
                }
        }
        dp(0);
        return f[0][k + 1] >= eps;
}

int main()
{
        
        read(k); read(n);
        for (int i = 1; i <= n; i++)
        {
                read(a[i]); 
                read(b[i]); 
                read(father[i]);
                addedge(father[i],i);
        }
        double l = 0 , r = 10000 , ans;
        while (l + eps < r)
        {
                double mid = (l + r) / 2.0;
                if (check(mid))
                {
                        l = mid;
                        ans = mid;
                } else r = mid;
        }
        printf("%.3lf
",ans);
        
        return 0;
    
}