uva11400

来源：互联网发布：5号电池多少毫安知乎编辑：程序博客网时间：2024/05/16 19:30

题目中要求为了省钱，可以把一些灯泡换成电压更高的另一些灯泡来节省电源的钱。于是，就有了按照电压排序后的动态规划

首先s[i]为前i种灯泡的数量之和，既L之和。然后定义 d[i]为1~i的最小花销，

那么d[i] = min(d[i], d[j] + (s[i]-s[j]) * c[i] + k[i]);(j < i).表示前j个用最佳方案，然后j+1 ~ i个全部都换成了电压更高的i号电源，因为电源相同的话就减少了点源的开销,答案为d[n];

/*********************************************** * Author: fisty * Created Time: 2015/2/12 17:17:46 * File Name   : uva11400.cpp *********************************************** */#include <iostream>#include <cstring>#include <deque>#include <cmath>#include <queue>#include <stack>#include <list>#include <map>#include <set>#include <string>#include <vector>#include <cstdio>#include <bitset>#include <algorithm>using namespace std;#define Debug(x) cout << #x << " " << x <<endl#define Memset(x, a) memset(x, a, sizeof(x))const int INF = 0x3f3f3f3f;typedef long long LL;typedef pair<int, int> P;#define FOR(i, a, b) for(int i = a;i < b; i++)#define MAX_N 1100int n;struct node{    int v;    int k;    int c;    int l;}lamp[MAX_N];bool cmp(node a, node b){    return a.v < b.v;}int dp[MAX_N];int s[MAX_N];int main() {    //freopen("in.cpp", "r", stdin);    //cin.tie(0);    //ios::sync_with_stdio(false);    while(cin >> n){           Memset(lamp, 0);        Memset(s, 0);        if(!n) break;        FOR(i, 1, n+1){            cin >> lamp[i].v >> lamp[i].k >> lamp[i].c >> lamp[i].l;        }        Memset(dp, 0);        sort(lamp + 1, lamp + 1 + n, cmp);        for(int i = 1;i <= n; i++){            s[i] = s[i-1] + lamp[i].l;        }        for(int i = 1;i <= n; i++){            for(int j = 0;j < i; j++){                if(j == 0){                    dp[i] = s[i]*lamp[i].c + lamp[i].k;                }else{                    dp[i] = min(dp[i],dp[j] + (s[i]-s[j])*lamp[i].c + lamp[i].k);                }            }        }        cout << dp[n] << endl;    }    return 0;}

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