zoj 2109 FatMouse' Trade 部分背包

       部分背包问题，贪心

       先按J[i] / F[i] 升序排序，每次选取J[i] / F[i]的最大值，如果不能取满J[i],则取把背包填满的容量即可。原因很简单，每次取J[i] / F[i]的最大值就使得背包单位体积价值最多。从大到小选择，自然会得到最优解。

 

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;

const int N = 1005;

struct info
{
	int j, f;
	double divid;
};
info a[N];

bool cmp(const info& a, const info& b)
{
	return a.divid > b.divid;
}
int main()
{
	int m, n;

	while (scanf("%d%d", &m, &n))
	{
		if (m == -1 && n == -1)
			break;

		for (int i = 1; i <= n; i++)
		{
			scanf("%d%d", &a[i].j, &a[i].f);
			a[i].divid = (double)a[i].j / a[i].f;
		}

		int beans = 0, catfood = 0;
		int i;
		sort(a+1, a+n+1, cmp);

		for (i = 1; i <= n; i++)
		{
			beans += a[i].j;
			catfood += a[i].f;
			if (catfood > m)
			{
				catfood -= a[i].f;
				beans -= a[i].j;
				break;
			}
		}

		if (i <= n)
			beans += (double)(m - catfood)/a[i].f * a[i].j;
		
		printf("%.3lf\n", beans);
	}
	return 0;
}