CF19B 01背包（必须装满）

http://codeforces.com/problemset/problem/19/B

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 ci and time ti 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 following n lines each item is described by a pair of numbers ti, ci(0 ≤ ti ≤ 2000, 1 ≤ ci ≤ 109). If ti 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 test(s)

input

42 100 201 51 3

output

8

input

30 10 100 100

output

111

/**cf19B题意：一个人想在超市偷东西...1秒偷一件，问偷过之后最后付的最少的钱是多少做法：就是买N件物品化最少的钱，以前是买一件物品可以得到一件，因为现在可以偷，可以认为是买一件物品送t件。背包必须完全装满，初始化注意*/#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;LL f[2005];int n;int main(){    while(~scanf("%d",&n))    {        LL x,y;        ///背包必须装满，初始化除了f[0]其余全为无穷        memset(f,0x3f3f3f3f3f,sizeof(f));        f[0]=0;        for(int i=1;i<=n;i++)        {            scanf("%I64d%I64d",&x,&y);            x++;            for(int j=n;j>=0;j--)            {                if(j>=x)                    f[j]=min(f[j],f[j-x]+y);                else                    f[j]=min(f[j],y);            }        }        printf("%I64d\n",f[n]);    }    return 0;}