POJ 1018(dp Or 枚举)


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

[Submit]   [Go Back]   [Status]  
[Discuss]

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1 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[i⑴][j]) {
for(int k=0;k<m;k++) {
int minb=min(b[k],j);
if(!dp[i][minb])dp[i][minb]=dp[i⑴][j]+p[k];
else dp[i][minb]=min(dp[i][minb],dp[i⑴][j]+p[k]);
}
}
}
double ans=0;
for(int i=1;i<=maxb;i++) {
if(!dp[n⑴][i])continue;
ans=max(ans,1.0*i/dp[n⑴][i]);
}
printf("%.3f
", ans);
}
return 0;
}

枚举竟然比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1 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=⑴;
for(int j=0;j<communication[i].size();++j) if(communication[i][j].first>=x) {
if(ch==⑴||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==⑴)return ⑴.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=⑶.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
", solve(n));
}
return 0;
}

波比源码 – 精品源码模版分享 | www.bobi11.com
1. 本站所有资源来源于用户上传和网络,如有侵权请邮件联系站长!
2. 分享目的仅供大家学习和交流,您必须在下载后24小时内删除!
3. 不得使用于非法商业用途,不得违反国家法律。否则后果自负!
4. 本站提供的源码、模板、插件等等其他资源,都不包含技术服务请大家谅解!
5. 如有链接无法下载、失效或广告,请联系管理员处理!
6. 本站资源售价只是赞助,收取费用仅维持本站的日常运营所需!
7. 本站源码并不保证全部能正常使用,仅供有技术基础的人学习研究,请谨慎下载
8. 如遇到加密压缩包,请使用WINRAR解压,如遇到无法解压的请联系管理员!

波比源码 » POJ 1018(dp Or 枚举)

发表评论

Hi, 如果你对这款模板有疑问,可以跟我联系哦!

联系站长
赞助VIP 享更多特权,建议使用 QQ 登录
喜欢我嘛?喜欢就按“ctrl+D”收藏我吧!♡