2005-2006 ACM-ICPC East Central North America Regional Contest (ECNA 2005) H.Two Ends

http://codeforces.com/gym/100650

分析：
比较水的题目。
给你一个序列，有两个人分别拿，每次只能拿最左或最右，其中一个是只会贪心的拿，问贪心的人最多能输几分。
写个DP搞两边就行了，给的序列是偶数个的（题意），所以2个2个搞，只有四种情况。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

typedef long long LL;
const double pi = acos(-1.0);
const long long mod = 1e9 + 7;
using namespace std;

int a[1005];
int dp[1005][1005];

int ans(int A,int B)
{
    if(dp[A][B] != 0)
        return dp[A][B];
    if(B - A == 1)
        return dp[A][B] = abs(a[A] - a[B]);
    int MAX1,MAX2;
    if(a[A + 1] >= a[B])
        MAX1 = ans(A + 2,B) + a[A] - a[A + 1];
    else
        MAX1 = ans(A + 1,B - 1) + a[A] - a[B];
    if(a[A] < a[B - 1])
        MAX2 = ans(A,B - 2) + a[B] - a[B - 1];
    else
        MAX2 = ans(A + 1,B - 1) + a[B] - a[A];
    return dp[A][B] = max(MAX1,MAX2);
}


int main()
{
    //freopen("int.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int N;
    int t = 0;
    while(scanf("%d",&N) && N)
    {
        for(int i = 1;i <= N;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        printf("In game %d, the greedy strategy might lose by as many as %d points.\n",++t,ans(1,N));
    }
    return 0;
}