poj1018 - Communication System

题目大意：有一个系统，这个系统由n种设备组成，现在每种设备有m个，设备有两种属性
一个是带宽，一个是价格
每种设备取一个，取满n个，怎么才能使所有带宽总和/总价格最大

解题思路：动态规划，dp[i][j]表示取到第i个设备带宽为j的最小的价格
那么有
for i = 1 : n
 forj = 0 : MAXV*MAXV
  for k = 1 : m
   dp[i][min(j,b[k]) = min(dp[i-1][j] + p[k] , dp[i][min(j,b[k]))

好像如果dp[i][j]表示取到第i个价格为j的最大带宽的话，空间会爆掉

 

#include <iostream>
#include <iomanip>
using namespace std;
#define MAXV 105
#define INF INT_MAX

int min(int a,int b){
	return a>b?b:a;
}

int max(int a,int b){
	return a<b?b:a;
}

class CDevice{
public:
	int m_nPrice;
	int m_nBandWidth;
};

class CComSystem{
public:
	void Run();
	CComSystem(){
		m_nDp = new int*[MAXV];
		for(int i = 0;i < MAXV;i++){
			m_nDp[i] = new int[MAXV*MAXV];

			for(int j = 0;j < MAXV*MAXV;j++){
				m_nDp[i][j] = INF;
			}
		}
		
		device = new CDevice[MAXV];
	}
	~CComSystem(){
		for(int i = 0;i < MAXV;i++){
			delete []m_nDp[i] ;
		}
		delete []m_nDp;
		delete []device;
	}
private:
	int **m_nDp;
	CDevice *device;
};

void CComSystem::Run(){
	int n, m ,maxBand = 0;
	int i, j ,k;
	double ans = -1;
	cin>>n;

	for(i = 0;i < n;i++){
		cin>>m;
		for(j = 0;j < m;j++){
			cin>>device[j].m_nBandWidth>>device[j].m_nPrice;
			maxBand = max(maxBand,device[j].m_nBandWidth);
		}

		if(i == 0){
			for(j = 0;j < m;j++){
				m_nDp[i][device[j].m_nBandWidth] 
					= min(device[j].m_nPrice,m_nDp[i][device[j].m_nBandWidth]);
			}
			continue;
		}

		for(j = 0;j <= maxBand;j++){
			if(m_nDp[i - 1][j]!=INF){
				for(k = 0;k < m;k++){
					m_nDp[i][min(j,device[k].m_nBandWidth)] = 
						min(m_nDp[i - 1][j] + device[k].m_nPrice, m_nDp[i][min(j,device[k].m_nBandWidth)]);
				}
			}
		}
	}

	for(j = 0;j < MAXV*MAXV;j++){
		if(m_nDp[n - 1][j]!=INF){
			if(ans < 1.0*j/m_nDp[n - 1][j]){
				ans = 1.0*j/m_nDp[n - 1][j];
			}
		}
	}

//	printf("%.3lf\n",ans);
	cout<<fixed<<setprecision(3)<<ans<<endl;  
}

int main(){
	int Case;
	cin>>Case;
	while(Case--){
		CComSystem com;
		com.Run();
	}
	return 0;
}