codeforces-148E-Porcelain【DP】

148E-Porcelain

                    time limit per test1 second     memory limit per test256 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 on n 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 of m shrieks can inflict on the collection of porcelain.

Input
The first line of input data contains two integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 10000). The next n lines contain the values of the items on the shelves: the first number gives the number of items on this shelf (an integer between 1 and 100, inclusive), followed by the values of the items (integers between 1 and 100, 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 least m.

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

input
2 3
3 3 7 2
3 4 1 5
output
15

input
1 3
4 4 3 1 2
output
9

题目链接：cf-148E

题目大意：有n个书架，每次取书只能从书架两端开始取，取m本书，问价值最大为多少

题目思路：今天比赛写了好久，可惜还差最后一步，少了一个for循环，最后发现已经没时间改了，再加上，我以为会超时orz结果并不会

①求出前缀和用zhi记录

②用sum数组存，sum[i][j]表示独立的第i个货架取j本书能达到的最大值

③dp[i][j]表示到第i个书架，取了j本书的最大值

以下是代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <algorithm>
#include <string>
#include <set>
#include <functional>
#include <numeric>
#include <sstream>
#include <stack>
#include <map>
#include <queue>
#include<iomanip>
using namespace std;
int dp[205][10005] = {0};
vector <int> ret[205];
int sum[205][205];
int zhi[205][205];
int main()
{
    int n,m;
    cin >> n >> m;
    for( int i=0 ; i<n ; i++ ){
        int t;
        cin >> t;
        for( int j=0 ; j<t ; j++ ){
            int num;
            cin >> num;
            ret[i].push_back(num);
            if( j==0 ) zhi[i][j] = ret[i][j];
            else zhi[i][j] = zhi[i][j-1]+ret[i][j];
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j <= ret[i].size(); j++)  //取j个 
        {
            int len = ret[i].size();
            for (int k = 0; k <= j ; k++){  //前面的长度 
                if(k == 0)
                {
                    sum[i][j]= max(sum[i][j],zhi[i][len - 1] - zhi[i][len - 1 - j]);
                } 
                else
                {

                    sum[i][j] = max(sum[i][j],zhi[i][k - 1] + zhi[i][len - 1] - zhi[i][len - 1 - j + k]);
                }
            }
        } 
    }
    int ans = 0;
    for (int i = 0; i <= m; i++)
    {
        if (i <= ret[0].size()) dp[0][i] = sum[0][i];
        else dp[0][i] = dp[0][ i - 1];
        ans = max(ans,dp[0][i]);
    }
    for (int i = 1; i < n; i++)  //前i个 
    {
        for (int j = 0; j <= m; j++)  
        {
            for (int k = 0; k <= ret[i].size() && k <= j; k++) //第i个取k个 
            {
                dp[i][j] = max(dp[i][j],dp[i - 1][j - k] + sum[i][k]);
                ans = max(ans,dp[i][j]);
            }
        } 
    }
    cout << ans << endl;
    return 0;
}
/* 2 10 3 4 5 6 5 1 2 3 4 5 */