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POJ 1018(dp Or 枚举）
Communication System
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;
}POJ 1018(dp Or 枚举）
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