[中等] UVa OJ 11400 Lighting system design 动态规划
2017-09-09 20:51
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题目描述
基本思路:(类比最长上升子序列LIS问题)
首先,对于每种电源的灯泡要么全换,要么不换,因为不能将大电压的灯泡换成小电压的,因此先将灯泡按电压大小排序,排完序后,就是一个考虑第i种灯泡换不换的问题了,是个多阶段决策问题,因此考虑动态规划。设d[i]为前i种灯泡的最小费用,s[i]为前i种灯泡的总数,c[i]为第i种灯泡的单价,k[i]为第i种灯泡的电源价格,则状态转移方程为:
d[i]=min{d[j]+(s[i]-s[j])*c[i]+k[i]},0=<j<=i,1<=i<=n
注意,上式中当j为0时表示将之前的所有灯泡全部换为第i种。边界条件d[0]=0
具体代码:
#include <iostream>
#include <algorithm>
using namespace std;
struct lamp
{
int v,k,c,l;
bool operator < (const lamp& rhs) const
{
return v<rhs.v;
}
};
const int maxn=1000+5;
lamp lp[maxn];
int s[maxn];
int d[maxn];
int main()
{
int n;
for(cin>>n;n!=0;cin>>n)
{
for(int i=1;i<=n;++i)
{
cin>>lp[i].v>>lp[i].k>>lp[i].c>>lp[i].l;
}
sort(lp+1,lp+1+n);
s[0]=0;
for(int i=1;i<=n;++i)
{
s[i]=s[i-1]+lp[i].l;
}
d[0]=0;
for(int i=1;i<=n;++i)
{
d[i]=d[i-1]+lp[i].l*lp[i].c+lp[i].k;
for(int j=0;j<i;++j)
d[i]=min(d[i],d[j]+(s[i]-s[j])*lp[i].c+lp[i].k);
}
cout<<d
<<endl;
}
return 0;
}
基本思路:(类比最长上升子序列LIS问题)
首先,对于每种电源的灯泡要么全换,要么不换,因为不能将大电压的灯泡换成小电压的,因此先将灯泡按电压大小排序,排完序后,就是一个考虑第i种灯泡换不换的问题了,是个多阶段决策问题,因此考虑动态规划。设d[i]为前i种灯泡的最小费用,s[i]为前i种灯泡的总数,c[i]为第i种灯泡的单价,k[i]为第i种灯泡的电源价格,则状态转移方程为:
d[i]=min{d[j]+(s[i]-s[j])*c[i]+k[i]},0=<j<=i,1<=i<=n
注意,上式中当j为0时表示将之前的所有灯泡全部换为第i种。边界条件d[0]=0
具体代码:
#include <iostream>
#include <algorithm>
using namespace std;
struct lamp
{
int v,k,c,l;
bool operator < (const lamp& rhs) const
{
return v<rhs.v;
}
};
const int maxn=1000+5;
lamp lp[maxn];
int s[maxn];
int d[maxn];
int main()
{
int n;
for(cin>>n;n!=0;cin>>n)
{
for(int i=1;i<=n;++i)
{
cin>>lp[i].v>>lp[i].k>>lp[i].c>>lp[i].l;
}
sort(lp+1,lp+1+n);
s[0]=0;
for(int i=1;i<=n;++i)
{
s[i]=s[i-1]+lp[i].l;
}
d[0]=0;
for(int i=1;i<=n;++i)
{
d[i]=d[i-1]+lp[i].l*lp[i].c+lp[i].k;
for(int j=0;j<i;++j)
d[i]=min(d[i],d[j]+(s[i]-s[j])*lp[i].c+lp[i].k);
}
cout<<d
<<endl;
}
return 0;
}
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