BZOJ5281 [Usaco2018 Open]Talent Show [0/1分数规划, 背包]

$Talent Show$

题目描述见链接 .

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$color{red}{正解部分}$

$0/1$ 分数规划,

二分出答案 $mid$, 要满足 $frac{sum t_i}{sum w_i} le mid ightarrow mid imes sum w_i - sum t_i geq 0 ightarrow sumleft( mid imes w_i - t_i ight) geq 0$ .

由于题目有总重量不小于 $W$ 的限制, 所以不能直接贪心选,

考虑 背包, 设 $F[i]$ 表示总重量为 $i$ 所能获得的最大值, 由于 $W$ 较小, 而 $w_i$ 比较大,
所以考虑将 $F[x geq W]$ 全部存在 $F[W]$ 中, 做 $0/1$ 背包 即可 .

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$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register

const int maxn = 1005;
const double inf = 1e10;
const double eps = 1e-9;

int N;
int W;
int w[maxn];
int t[maxn];

double F[maxn];

bool check(double mid){
        for(reg int i = 1; i <= W; i ++) F[i] = -inf;
        F[0] = 0;
        for(reg int i = 1; i <= N; i ++)
                for(reg int j = W; j >= 0; j --){
                        double &tmp = F[std::min(W, j+w[i])];
                        tmp = std::max(tmp, F[j] + 1.0*t[i] - 1.0*mid*w[i]);
                }
        return F[W] >= 0;
}

int main(){
        scanf("%d%d", &N, &W);
        double l = 0, r = 0;
        for(reg int i = 1; i <= N; i ++) scanf("%d%d", &w[i], &t[i]), r += t[i];
        double Ans = 0;
        while(r-l > eps){
                double mid = (l+r)/2;
                if(check(mid)) Ans = std::max(Ans, mid), l = mid + eps;
                else r = mid - eps;
        }
        printf("%d
", (int)(Ans*1000));
        return 0;
}