uva 11400——Lighting System Design
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题意: 给定一些灯泡,每种灯泡有不同的电压v,电源费用k,每个灯泡的费用c,所需的灯泡的数量L,同种灯泡可以使用相同电源,问最小花费。
思路:dp,每种灯泡要么选要么不选两种状态,状态转移方程是d(i)=min(d[j]+(s[i]-s[j])*c[i]+k[i])。d[i]保存的是最小的开销。
code:
#include <bits/stdc++.h>using namespace std;#define ft(i,s,t) for (int i=s;i<=t;i++)const int N=1005;const int INF=0x3f3f3f3f;struct node{ int v,k,c,l;}g[N];int d[N],s[N];bool cmp(node A,node B){ return A.v<B.v;}int main(){ int n; while (~scanf("%d",&n),n) { ft(i,1,n) { scanf("%d%d%d%d",&g[i].v,&g[i].k,&g[i].c,&g[i].l); } sort(g+1,g+1+n,cmp);s[0]=0; ft(i,1,n) s[i]=s[i-1]+g[i].l; ft(i,1,n) { d[i]=s[i]*g[i].c+g[i].k; ft(j,1,i) d[i]=min(d[i],d[j]+(s[i]-s[j])*g[i].c+g[i].k); } printf("%d\n",d[n]); }}
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