CodeForces 148E Porcelain

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题意：第一行给出两个数n,m，n表示下面有n行数据，每行数据可以从前或从后去若干个数，一共取m个数，求取出的n个数和最大值

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

思路：预处理背包，将每行数据预处理，求出每行处理成max_n = ( left_sum_n, right_sum_n, left_right_sum_n )，后面按正常背包处理即可

注意点：无

以下为AC代码：

#AuthorProblemLangVerdictTimeMemorySentJudged9742020Practice:
luminous11148E -12GNU C++11Accepted278 ms796 KB2015-02-07 07:59:222015-02-07 07:59:22

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <vector>#include <deque>#include <list>#include <cctype>#include <algorithm>#include <climits>#include <queue>#include <stack>#include <cmath>#include <map>#include <set>#include <iomanip>#include <cstdlib>#include <ctime>#define ll long long#define ull unsigned long long#define all(x) (x).begin(), (x).end()#define clr(a, v) memset( a , v , sizeof(a) )#define pb push_back#define mp make_pair#define read(f) freopen(f, "r", stdin)#define write(f) freopen(f, "w", stdout)using namespace std;const double pi = acos(-1);int tmp[100005] = { 0 };int dp[100005] = { 0 };int num[105] = { 0 };int ls[105] = { 0 };int rs[105] = { 0 };int ss[105] = { 0 };int main(){    ios::sync_with_stdio( false );    int m, n;    while ( cin >> n >> m ){        int k;        clr ( dp, 0 );        clr ( num, 0 );        for ( int i = 0; i < n; i ++ ){            cin >> k;            for ( int j = 1; j <= k; j ++ ){                cin >> num[j];            }            clr ( ls, 0 );            clr ( rs, 0 );            clr ( ss, 0 );            for ( int p = 1; p <= k; p ++ ){                ls[p] = ls[p-1] + num[p];            }            for ( int p = k , p1 = 1; p > 0; p --, p1 ++ ){                rs[p1] = rs[p1-1] + num[p];            }            for ( int p = 1; p <= k; p ++ ){                ss[p] = max ( ss[p], rs[p] );                ss[p] = max ( ss[p], ls[p] );            }            for ( int a = 1; a <= k; a ++ ){                for ( int b = 1; b <= k; b ++ ){                    if ( a + b <= k ){                        ss[a+b] = max ( ss[a+b], ls[a] + rs[b] );                    }                }            }            clr ( tmp, 0 );            memcpy ( tmp, dp, sizeof( dp ) );            for ( int a = 1; a <= k; a ++ ){                for ( int p = m; p >= a; p -- ){                    dp[p] = max ( dp[p], tmp[p-a] + ss[a] );                }            }        }        cout << dp[m] << endl;    }    return 0;}

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