【DP】HDU-1300 Pearls

Pearls

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Problem Description

In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name because it delivers to the royal family of Pearlania. But it also produces bracelets and necklaces for ordinary people. Of course the quality of the pearls for these people is much lower then the quality of pearls for the royal family. In Pearlania pearls are separated into 100 different quality classes. A quality class is identified by the price for one single pearl in that quality class. This price is unique for that quality class and the price is always higher then the price for a pearl in a lower quality class.

Every month the stock manager of The Royal Pearl prepares a list with the number of pearls needed in each quality class. The pearls are bought on the local pearl market. Each quality class has its own price per pearl, but for every complete deal in a certain quality class one has to pay an extra amount of money equal to ten pearls in that class. This is to prevent tourists from buying just one pearl.

Also The Royal Pearl is suffering from the slow-down of the global economy. Therefore the company needs to be more efficient. The CFO (chief financial officer) has discovered that he can sometimes save money by buying pearls in a higher quality class than is actually needed. No customer will blame The Royal Pearl for putting better pearls in the bracelets, as long as the prices remain the same.

For example 5 pearls are needed in the 10 Euro category and 100 pearls are needed in the 20 Euro category. That will normally cost: (5+10)*10 + (100+10)*20 = 2350 Euro.

Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.

The problem is that it requires a lot of computing work before the CFO knows how many pearls can best be bought in a higher quality class. You are asked to help The Royal Pearl with a computer program.

Given a list with the number of pearls and the price per pearl in different quality classes, give the lowest possible price needed to buy everything on the list. Pearls can be bought in the requested, or in a higher quality class, but not in a lower one.

 

Input

The first line of the input contains the number of test cases. Each test case starts with a line containing the number of categories c (1 <= c <= 100). Then, c lines follow, each with two numbers ai and pi. The first of these numbers is the number of pearls ai needed in a class (1 <= ai <= 1000). The second number is the price per pearl pi in that class (1 <= pi <= 1000). The qualities of the classes (and so the prices) are given in ascending order. All numbers in the input are integers.

 

Output

For each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.

 

Sample Input

2
2
100 1
100 2
3
1 10
1 11
100 12

 

Sample Output

330
1344

 

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题意：给出n种珠宝，每种价格不同，有一定的需求量。若要购买某种珠宝，必须先付10单位的该珠宝价格。按珠宝价格从小到大输入。可以用更高价格的珠宝来代替低价格的珠宝，但是反之不行。问最后的最小花费。

前言：之前自己独立想了一个。但是没有紧扣题意。需要三重循环。复杂度危险。

思路：

首先要明白，最高价格的那种珠宝是必须要买的，因此设dp[i]表示考虑到第i种珠宝时的最小花费。

1. 考虑1种珠宝的情况：必须买

2. 考虑2种珠宝的情况：可以从买了第1种珠宝的情况转移过来，也就是说不替换，直接买了第2种；也可以不买第1种，而是从买了第0种（设为不存在）珠宝的情况转移过来。这样的话，购买量是第1种的需求量加上第二种。两种决策取最优。

3. 考虑3种珠宝的情况：要么1和2全换成3，要么2全换成3，要么全部不换。三种决策取最优。

状态转移方程为：dp[i] = min{dp[0...i-1] + (sum of need[j...i] + 10) * p[i]}

利用前缀和O(1)计算需求量之和。

P.S. 

1. DP需要仔细分析边界情况，不仅是0，还有n。对于边界情况常常隐含着某种对于状态的限制。

2. 这种推理性问题（一眼下来我们并不知道用哪些珠宝来代替哪些珠宝）往往需要模拟一下0、1、2这些初始情况，从中得到转移方式。

另外给出我错误的思路。。dp[i][j]表示前i种珠宝，购买了j个。对于这种方式显然没有正确描述，如果要用更高价格代替某一价格，那么该价格所有珠宝都要代替。不合题意。

代码如下：

/**
 * ID: j.sure.1
 * PROG:
 * LANG: C++
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <climits>
#include <iostream>
#define For(i, x, y) for(int i=x; i<y; i++)
#define For_(i, x, y) for(int i=x; i>=y; i--)
#define Mem(f, x) memset(f, x, sizeof(f))
#define Sca(x) scanf("%d", &x)
#define Pri(x) printf("%d\n", x)
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
/****************************************/
const int N = 111;
int n, need[N], pri[N], dp[N];

int main()
{
#ifdef J_Sure
    freopen("000.in", "r", stdin);
    freopen("999.out", "w", stdout);
#endif
    int T;
    Sca(T);
    while(T--) {
        Sca(n);
        need[0] = 0;
        For(i, 1, n+1) {
            scanf("%d%d", &need[i], &pri[i]);
            need[i] += need[i-1];
        }
        dp[0] = 0;
        For(i, 1, n+1) {
            int ret = INF;
            For(j, 0, i) {
                //将j+1~i的珠宝全部换成i
                ret = min(ret, dp[j] + (need[i]-need[j]+10) * pri[i]);
            }
            dp[i] = ret;
        }
        printf("%d\n", dp[n]);
    }
    return 0;
}