P4377

发表于 | 阅读数

求选中最大

并且$\sum b[i]\geq w$

01分数规划模版题。

考虑$01$规划，二分。$\sum \frac{b[i]}{a[i]}\geq ans\rightarrow \sum {b[i]}-mid\times {a[i] }\geq 0$

设$c[i]=\sum {b[i]}-mid\times {a[i] }$，做个背包就可以判断第二个条件

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const double eps = 1e-6;
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}
double dp[N], c[N];
int a[N], b[N];
int n, W;
bool check(double x)
{
    for (int i = 1; i <= n; i++)
        c[i] = double(b[i]) - x * double(a[i]);

    for (int i = 0; i <= W; i++)
        dp[i] = -1e9;
    dp[0] = 0;
    for (int i = 1; i <= n; i++)
    {
        for (int j = W; j >= 0; j--)
        {
            if (j + a[i] >= W)
                dp[W] = max(dp[W], dp[j] + c[i]);
            else
                dp[j + a[i]] = max(dp[j + a[i]], dp[j] + c[i]);
        }
    }

    return dp[W] >= 0;
}
int main()
{
    n = read(), W = read();
    for (int i = 1; i <= n; i++)
        a[i] = read(), b[i] = read();
    double l = 0, r = 1e6;
    while (l + eps < r)
    {
        double mid = (l + r) / 2;
        if (check(mid))
            l = mid;
        else
            r = mid;
    }

    cout << (int)(l * 1000) << endl;
}