CF-19B - Checkout Assistant(DP)

B - Checkout Assistant

Crawling in process...Crawling failedTime Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

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Description

Bob came to a cash & carry store, put n items into his trolley, and went to the checkout counter to pay. Each item is described by its price c_i and time t_i

Input

The first input line contains number n (1 ≤ n ≤ 2000). In each of the following n lines each item is described by a pair of numbers t_i, c_i (0 ≤ t_i ≤ 2000, 1 ≤ c_i ≤ 10⁹). If t_i is 0, Bob won't be able to steal anything, while the checkout assistant is occupied with item i.

Output

Output one number — answer to the problem: what is the minimum amount of money that Bob will have to pay.

Sample Input

Input

4
2 10
0 20
1 5
1 3

Output

8

Input

3
0 1
0 10
0 100

Output

111

思路：简单01背包，不过要注意优化，就是题目中有说偷的时间只要1秒，所以当时间大于n秒时，他可以都所以物品，因此只要更新到 n就可以得到正确答案了。

 

仔细想想，记忆化搜索好像也行。

#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
using namespace std;
const long long oo=1e18;
const int mm=2005;
long long f[mm];
int m,x;
long long y,z;
int main()
{
  while(cin>>m)
  {
    f[0]=0;
    for(int i=1;i<mm;i++)
      f[i]=oo;
    for(int i=0;i<m;i++)
    {cin>>x>>y;
      int mid;
      for(int j=m;j>=0;j--)
      {
        if(j+x+1<m)mid=j+x+1;///时间大于m时，他可以偷所有物品因此只需到m就行
        else mid=m;
        if(f[j]+y<f[mid])///更新值
          f[mid]=f[j]+y;
      }
    }
    cout<<f[m]<<"\n";
  }
}