CF-19B - Checkout Assistant(DP)

B - Checkout Assistant

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

SubmitStatusPracticeCodeForces 19B

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 pricec_i and timet_i in seconds that a checkout assistant spends on this item. While the checkout assistant is occupied with some item, Bob can steal some other items from his trolley. To steal one item Bob needs exactly 1 second. What is the minimum amount of money that Bob will have to pay to the checkout assistant? Remember, please, that it is Bob, who determines the order of items for the checkout assistant.

Input

The first input line contains number n (1 ≤ n ≤ 2000). In each of the followingn lines each item is described by a pair of numberst_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 itemi.

Output

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

Sample Input

Input

42 100 201 51 3

Output

8

Input

30 10 100 100

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";  }}