POJ 1018(dp Or 枚举）

最新推荐文章于 2023-08-08 14:59:34 发布

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最新推荐文章于 2023-08-08 14:59:34 发布

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分类专栏： dp 文章标签：动态规划枚举

本文链接：https://blog.youkuaiyun.com/acvcla/article/details/42302047

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本文介绍了一个特殊的通信系统设计问题，目标是在多个设备中选择最佳制造商组合以最大化整体带宽与价格的比例。通过动态规划和枚举策略实现了这一目标，并提供了详细的代码实现。

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Communication System

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23738		Accepted: 8437

Description

We have received an order from Pizoor Communications Inc. for a special communication system. The system consists of several devices. For each device, we are free to choose from several manufacturers. Same devices from two manufacturers differ in their maximum bandwidths and prices.
By overall bandwidth (B) we mean the minimum of the bandwidths of the chosen devices in the communication system and the total price (P) is the sum of the prices of all chosen devices. Our goal is to choose a manufacturer for each device to maximize B/P.

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case starts with a line containing a single integer n (1 ≤ n ≤ 100), the number of devices in the communication system, followed by n lines in the following format: the i-th line (1 ≤ i ≤ n) starts with mi (1 ≤ mi ≤ 100), the number of manufacturers for the i-th device, followed by mi pairs of positive integers in the same line, each indicating the bandwidth and the price of the device respectively, corresponding to a manufacturer.

Output

Your program should produce a single line for each test case containing a single number which is the maximum possible B/P for the test case. Round the numbers in the output to 3 digits after decimal point.

Sample Input

1 3
3 100 25 150 35 80 25
2 120 80 155 40
2 100 100 120 110

Sample Output

0.649

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

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dp做法，dp[i][j]表示前i个物品最小带宽为j所需的最小费用。

/************************************************************************* 
    > File Name: xiaozhao.cpp 
    > Author: acvcla 
    > QQ:acvcla@gmail.com  
    > Mail: acvcla@gmail.com
    > Created Time: 2014年12月27日 星期一 22时34分13秒 
*************************************************************************/  
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int maxn=1e2+10;
const int maxv=1e3+10;
int dp[maxn][maxv],b[maxn],p[maxn];
int main()
{
	int T,n,m;
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&n);
		memset(dp,0,sizeof dp);
		int maxb=0;

		for(int i=0;i<n;++i) {
			scanf("%d",&m);

			for(int j=0;j<m;++j){
				scanf("%d%d",&b[j],&p[j]);
				maxb=max(maxb,b[j]);	
			}
			if(i==0) {
					for(int j=0;j<m;++j) {
						if(!dp[0][b[j]])dp[0][b[j]]=p[j];
						else dp[0][b[j]]=min(dp[0][b[j]],p[j]);
					}
					continue;
			}
			for(int j=1;j<=maxb;j++) if(dp[i-1][j]) {
				for(int k=0;k<m;k++) {
					int minb=min(b[k],j);
					if(!dp[i][minb])dp[i][minb]=dp[i-1][j]+p[k];
					else dp[i][minb]=min(dp[i][minb],dp[i-1][j]+p[k]);
				}
			}
		}
		double ans=0;
		for(int i=1;i<=maxb;i++) {
			if(!dp[n-1][i])continue;
			ans=max(ans,1.0*i/dp[n-1][i]);
		}
		printf("%.3f\n", ans);
	}
	return 0;
}

枚举居然比dp更快。。。，枚举最小带宽。

/************************************************************************* 
    > File Name: xiaozhao.cpp 
    > Author: acvcla 
    > QQ:acvcla@gmail.com  
    > Mail: acvcla@gmail.com
    > Created Time: 2014年12月27日 星期一 22时34分13秒 
*************************************************************************/  
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
typedef long long ll;
typedef pair<int,int>pii;
const int maxn=1e4+10;
const int inf=0x3f3f3f3f;
vector<pii>communication[maxn];
vector<int> v;
double work(int x,int n)
{
	int b=inf;
	double p=0;
	for(int i=0;i<n;i++) {
		int ch=-1;
		for(int j=0;j<communication[i].size();++j) if(communication[i][j].first>=x) {
			if(ch==-1||communication[i][j].second<communication[i][ch].second||
				(communication[i][j].second==communication[i][ch].second&&communication[i][j].first>communication[i][ch].first)){

				ch=j;
			}
		}
		if(ch==-1)return -1.0;
		b=min(b,communication[i][ch].first);
		p+=communication[i][ch].second;
	}
	return b/p;
}
double solve(int n)
{
	sort(v.begin(), v.end());
	unique(v.begin(), v.end());

	int L=0,R=v.size();
	double ans=-3.0,p=0;
	for(int i=0;i<v.size();++i) {
		p=work(v[i],n);
		if(p<=0)return ans;
		ans=max(p,ans);
	}
	return ans;
}
int main()
{
	int T,n,m;
	scanf("%d",&T);
	while(T--) {
		scanf("%d",&n);

		for(int i=0;i<n;i++)communication[i].clear();
		v.clear();

		for(int i=0;i<n;i++) {
			scanf("%d",&m);
			int b,p;

			while(m--) {
				scanf("%d%d",&b,&p);
				v.push_back(b);
				communication[i].push_back(make_pair(b,p));
			}
		}
		printf("%.3f\n", solve(n));
	}
	return 0;
}