POJ 3186 Treats for the Cows

題意：有一個序列num，只能從兩端取出，第i個取出的數的代價爲i*num(i)，求將這個序列中所有數全部取出的最大代價

鏈接：http://poj.org/problem?id=3186

思路：區間dp，利用記憶化搜索查找[l,r]中的合併的最大值，記錄在dp[l][r]中

注意點：無

以下爲AC代碼：

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
14186637	luminous11	3186	Accepted	16504K	125MS	G++	2989B	2015-05-12 15:32:33

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <cctype>
#include <climits>
#include <iomanip>
#include <cstdlib>
#include <algorithm>
//#include <unordered_map>
//#include <unordered_set>
#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 RDI(a) scanf ( "%d", &a )
#define RDII(a, b) scanf ( "%d%d", &a, &b )
#define RDIII(a, b, c) scanf ( "%d%d%d", &a, &b, &c )
#define RS(s) scanf ( "%s", s )
#define PI(a) printf ( "%d", a )
#define PIL(a) printf ( "%d\n", a )
#define PII(a,b) printf ( "%d %d", a, b )
#define PIIL(a,b) printf ( "%d %d\n", a, b )
#define PIII(a,b,c) printf ( "%d %d %d", a, b, c )
#define PIIIL(a,b,c) printf ( "%d %d %d\n", a, b, c )
#define PL() printf ( "\n" )
#define PSL(s) printf ( "%s\n", s )
#define rep(i,m,n) for ( int i = m; i <  n; i ++ )
#define REP(i,m,n) for ( int i = m; i <= n; i ++ )
#define dep(i,m,n) for ( int i = m; i >  n; i -- )
#define DEP(i,m,n) for ( int i = m; i >= n; i -- )
#define repi(i,m,n,k) for ( int i = m; i <  n; i += k )
#define REPI(i,m,n,k) for ( int i = m; i <= n; i += k )
#define depi(i,m,n,k) for ( int i = m; i >  n; i += k )
#define DEPI(i,m,n,k) for ( int i = m; i >= n; i -= k )
#define READ(f) freopen(f, "r", stdin)
#define WRITE(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
template <class T>
inline bool RD ( T &ret )
{
    char c;
    int sgn;
    if ( c = getchar(), c ==EOF )return 0; //EOF
    while ( c != '-' && ( c < '0' || c > '9' ) ) c = getchar();
    sgn = ( c == '-' ) ? -1 : 1;
    ret = ( c == '-' ) ? 0 : ( c - '0' );
    while ( c = getchar() , c >= '0' && c <= '9' ) ret = ret * 10 + ( c - '0' );
    ret *= sgn;
    return 1;
}
inline void PD ( int x )
{
    if ( x > 9 ) PD ( x / 10 );
    putchar ( x % 10 + '0' );
}
const double eps = 1e-10;
const int dir[4][2] = { 1,0, -1,0, 0,1, 0,-1 };
struct node{
    int x, y, cnt;
    node(){}
    node( int _x, int _y ) : x(_x), y(_y) {}
    node( int _x, int _y, int _cnt ) : x(_x), y(_y), cnt(_cnt) {}
};
int num[2005];
int dp[2005][2005];
int dfs ( int l, int r, int k )
{
    if ( l > r ) return 0;
    if ( dp[l][r] ) return dp[l][r];
    int m = dfs ( l + 1, r, k + 1 ) + k * num[l];
    int n = dfs ( l, r - 1, k + 1 ) + k * num[r];
    dp[l][r] = max ( m, n );
    return dp[l][r];
}
int main()
{
    int n;
    //READ ( "data.in" );
    while ( RDI ( n ) != EOF ){
        clr ( dp, 0 );
        rep ( i, 0, n ){
            RDI ( num[i] );
        }
        int ans = dfs ( 0, n - 1, 1 );
        PIL ( ans );
    }
    return 0;
}