题意：共有n种(n<=1000)种灯泡，每种灯泡用4个数值表示。电压V(V<=132000)，电源费用K(K<=1000)，每个灯泡的费用C(C<=10)和所需灯泡的数量L(1<=L<=100)。把一些灯泡换成电压更高的另一种灯泡以节省电源的钱（不能换成电压更低的灯泡）。计算出最优方案费用。
分析：
1、每种电压的灯泡要么全换要么全不换，否则电压不同，需要买更多电源不划算。
2、把灯泡按电压从小到大排序。
3、sum[i]---前i种灯泡的总数量
4、dp[i]----灯泡1~i的最小开销
5、状态转移方程dp[i] = Min(dp[i], dp[j] + (sum[i] - sum[j]) * num[i].c + num[i].k);表示前j个先用最优方案买，第j+1~i个用第i号的电源。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int sum[MAXN];
int dp[MAXN];
struct Node{
int v, k, c, l;
void read(){
scanf("%d%d%d%d", &v, &k, &c, &l);
}
bool operator < (const Node& rhs)const{
return v < rhs.v;
}
}num[MAXN];
int main(){
int n;
while(scanf("%d", &n) == 1){
if(!n) return 0;
for(int i = 1; i <= n; ++i){
num[i].read();
}
sort(num + 1, num + n + 1);
sum[0] = 0;
for(int i = 1; i <= n; ++i){
sum[i] = sum[i - 1] + num[i].l;
}
dp[0] = 0;
for(int i = 1; i <= n; ++i){
dp[i] = INT_INF;
for(int j = 0; j < i; ++j){
dp[i] = Min(dp[i], dp[j] + (sum[i] - sum[j]) * num[i].c + num[i].k);
}
}
printf("%d\n", dp[n]);
}
return 0;
}