Codeforces 148E Porcelain (预处理＋多重背包)

E. Porcelain

time limit per test：3 seconds

memory limit per test：256 megabytes

During her tantrums the princess usually smashes some collectable porcelain. Every furious shriek is accompanied with one item smashed.

The collection of porcelain is arranged neatly onn shelves. Within each shelf the items are placed in one row, so that one can access only the outermost items — the leftmost or the rightmost item, not the ones in the middle of the shelf. Once an item is taken, the next item on that side of the shelf can be accessed (see example). Once an item is taken, it can't be returned to the shelves.

You are given the values of all items. Your task is to find the maximal damage the princess' tantrum ofm shrieks can inflict on the collection of porcelain.

Input

The first line of input data contains two integersn (1 ≤ n ≤ 100) andm (1 ≤ m ≤ 10000). The nextn lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between1 and100, inclusive), followed by the values of the items (integers between1 and100, inclusive), in the order in which they appear on the shelf (the first number corresponds to the leftmost item, the last one — to the rightmost one). The total number of items is guaranteed to be at leastm.

Output

Output the maximal total value of a tantrum of m shrieks.

Sample test(s)

Input

2 3
3 3 7 2
3 4 1 5

Output

15

Input

1 3
4 4 3 1 2

Output

9

Note

In the first case there are two shelves, each with three items. To maximize the total value of the items chosen, one can take two items from the left side of the first shelf and one item from the right side of the second shelf.

In the second case there is only one shelf, so all three items are taken from it — two from the left side and one from the right side.

题目链接：http://codeforces.com/contest/148/problem/e

题目大意：给n层数字，一共要取m次，每次取数字只能从任意一层的最左或最右取

题目分析：预处理出每层从两边取j (j <= m)个的所有可能的最大值，之后就是个普通的多重背包问题

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int sum[105], ma[105], dp[10005];

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i++)
    {
        int num;
        scanf("%d", &num);
        for(int j = 1; j <= num; j++)
        {
            int tmp;
            scanf("%d", &tmp);
            sum[j] = sum[j - 1] + tmp;  //计算前缀和
        }
        //预处理
        memset(ma, 0, sizeof(ma));
        for(int j = 0; j <= num; j++)
            for(int k = 0; k <= j; k++)
                ma[j] = max(ma[j], sum[k] + sum[num] - sum[num + k - j]);
        printf("\n");
        //多重背包
        for(int j = m; j >= 1; j--)
            for(int k = 1; k <= min(j, num); k ++)
                dp[j] = max(dp[j], dp[j - k] + ma[k]);
    }
    printf("%d\n", dp[m]);
}