Aragorn's Story

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/148/E

Description

Our protagonist is the handsome human prince Aragorn comes from The Lord of the Rings. One day Aragorn finds a lot of enemies who want to invade his kingdom. As Aragorn knows, the enemy has N camps out of his kingdom and M edges connect them. It is guaranteed that for any two camps, there is one and only one path connect them. At first Aragorn know the number of enemies in every camp. But the enemy is cunning , they will increase or decrease the number of soldiers in camps. Every time the enemy change the number of soldiers, they will set two camps C1 and C2. Then, for C1, C2 and all camps on the path from C1 to C2, they will increase or decrease K soldiers to these camps. Now Aragorn wants to know the number of soldiers in some particular camps real-time. A的某一段完全重合，或者能够经过上下左右平移与折线A的某一段完全重合，则表示秋实大哥吹出了妹子的一部分旋律。

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.

 

Sample Input

2 3
3 3 7 2
3 4 1 5

Sample Output

15

HINT

 

题意

有n排花盆，每排有k个，然后有个人想扔m个花瓶，每个花瓶有个价值val

他只能选择每一排的最左边或者最右边扔

求扔的最大价值

题解：

背包问题，bag[i][j]表示第i排扔j个的最大值

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int bag[1000][1000];
int t[1000];
int a[1000][1000];
int dp[105][10001];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&t[i]);
        for(int j=1;j<=t[i];j++)
        {
            scanf("%d",&a[i][j]);
            a[i][j]+=a[i][j-1];
        }
    }
    for(int i=1;i<=n;i++)
        for(int j=1;j<=t[i];j++)
            for(int k=0;k<=j;k++)
                bag[i][j]=max(bag[i][j],a[i][k]+a[i][t[i]]-a[i][t[i]-j+k]);
    for(int i=1;i<=n;i++)
    {
        for(int j=m;j>=0;j--)
        {
            for(int k=0;k<=t[i]&&k<=j;k++)
            {
                dp[i][j]=max(dp[i][j],dp[i-1][j-k]+bag[i][k]);
            }
        }
    }
    cout<<dp[n][m]<<endl;

}