Lighting System Design

题意：从小到大给出额定功率，给出该功率费用，和灯泡的数量和单价，现在灯泡能在比他额定功率大的功率运行，求让所有灯泡正常工作的最小费用

分析：

问题转化为求用哪几个功率运行灯泡最小费用，dp[i]前i个功率的灯泡正常最小费用dp[i]=min(dp[i],dp[j]+num)(j<i);

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cctype>
#include <complex>
#include <cassert>
#include <utility>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;
#define lson l,m,rt<<1
#define pi acos(-1.0)
#define rson m+1,r,rt<<11
#define All 1,N,1
#define read freopen("in.txt", "r", stdin)
const ll  INFll = 0x3f3f3f3f3f3f3f3fLL;
const int INF= 0x7ffffff;
const int mod =  1000000007;
struct node{
    int p,k,c,l;
}d[1010];
int sum[1010];
bool cmp(node x,node y){
    return  x.p<y.p;
}
int dp[1010],n;
void solve(){
    sort(d+1,d+n+1,cmp);
    memset(sum,0,sizeof(sum));
    sum[0]=0;
    for(int i=1;i<=n;++i)
        sum[i]=sum[i-1]+d[i].l;
    for(int i=1;i<=n;++i)
        dp[i]=d[i].c*sum[i]+d[i].k;
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<i;++j)
            dp[i]=min(dp[i],dp[j]+(sum[i]-sum[j])*d[i].c+d[i].k);
    }
    printf("%d\n",dp[n]);
}
int main()
{
    while(~scanf("%d",&n)){
        if(n==0)break;
        for(int i=1;i<=n;++i){
            scanf("%d%d%d%d",&d[i].p,&d[i].k,&d[i].c,&d[i].l);
        }
        solve();
    }
return 0;
}