POJ 2738 Two Ends(记忆化）

Description

In the two-player game "Two Ends", an even number of cards is laid out in a row. On each card, face up, is written a positive integer. Players take turns removing a card from either end of the row and placing the card in their pile. The player whose cards add
up to the highest number wins the game. Now one strategy is to simply pick the card at the end that is the largest -- we'll call this the greedy strategy. However, this is not always optimal, as the following example shows: (The first player would win if she
would first pick the 3 instead of the 4.) 

3 2 10 4 

You are to determine exactly how bad the greedy strategy is for different games when the second player uses it but the first player is free to use any strategy she wishes.

Input

There will be multiple test cases. Each test case will be contained on one line. Each line will start with an even integer n followed by n positive integers. A value of n = 0 indicates end of input. You may assume that n is no more than 1000. Furthermore, you
may assume that the sum of the numbers in the list does not exceed 1,000,000.

Output

For each test case you should print one line of output of the form: 

In game m, the greedy strategy might lose by as many as p points. 

where m is the number of the game (starting at game 1) and p is the maximum possible difference between the first player's score and second player's score when the second player uses the greedy strategy. When employing the greedy strategy, always take the larger
end. If there is a tie, remove the left end.

Sample Input

4 3 2 10 4
8 1 2 3 4 5 6 7 8
8 2 2 1 5 3 8 7 3
0

Sample Output

In game 1, the greedy strategy might lose by as many as 7 points.
In game 2, the greedy strategy might lose by as many as 4 points.
In game 3, the greedy strategy might lose by as many as 5 points.

Source

East Central North America 2005

上次比赛写了个记忆化，这次算是跪了。參照：点击打开链接

题意：第二个人是贪心拿法，仅仅能从两端拿，问最大的差值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
using namespace std;
const int maxn=1100;
int dp[maxn][maxn];
int a[maxn],n;
int solve(int x,int y)
{
   if(dp[x][y]!=-1)
      return dp[x][y];
   if(x+1==y)
      return dp[x][y]=abs(a[x]-a[y]);
   int sa,sb;
   if(a[x+1]>=a[y])//第一个人选左端
      sa=solve(x+2,y)+a[x]-a[x+1];
   else
      sa=solve(x+1,y-1)+a[x]-a[y];
   if(a[x]<a[y-1])//第一个人选右端
      sb=solve(x,y-2)+a[y]-a[y-1];
   else
      sb=solve(x+1,y-1)+a[y]-a[x];
   return dp[x][y]=max(sa,sb);
}

int main()
{
    int l=0;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        memset(dp,-1,sizeof(dp));
        printf("In game %d, the greedy strategy might lose by as many as %d points.\n",++l,solve(0,n-1));
    }
    return 0;
}