Game of Sum - UVa 10891 dp

Problem E

Game of Sum

Input File: e.in
Output: Standard Output

This is a two player game. Initially there are n integer numbers in an array and players A and B get chance to take them alternatively. Each player can take one or more numbers from the left or
right end of the array but cannot take from both ends at a time. He can take as many consecutive numbers as he wants during his time. The game ends when all numbers are taken from the array by the players. The point of each player is calculated by the summation
of the numbers, which he has taken. Each player tries to achieve more points from other. If both players play optimally and player A starts the game then how much more point can player A get than player B?

Input

The input consists of a number of cases. Each case starts with a line specifying the integer n (0 < n ≤100), the number of elements in the array. After that, n numbers are given for the game. Input is terminated
by a line where n=0.



Output

For each test case, print a number, which represents the maximum difference that the first player obtained after playing this game optimally.

Sample Input Output for Sample Input

4 4 -10 -20 7 4 1 2 3 4 0

7 10

题意：一个人可以从左边或右边取任意个连续的数，得到其和的积分，问先手最多比后手多多少分。

思路：dp[i][j][0]表示在i到j的区间内先手从左边开始取的最大分差，dp[i][j][1]表示在i到j的区间内先手从右边开始取的最大分差，具体转移方程看代码吧，也不是很长。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int num[110],dp[110][110][2];
int main()
{ int n,i,j,k;
  while(~scanf("%d",&n) && n)
  { for(i=1;i<=n;i++)
    { scanf("%d",&num[i]);
      dp[i][i][0]=dp[i][i][1]=num[i];
    }
    for(i=n-1;i>=1;i--)
     for(j=i+1;j<=n;j++)
     { dp[i][j][0]=max(num[i]+dp[i+1][j][0],num[i]-max(dp[i+1][j][0],dp[i+1][j][1]));
       dp[i][j][1]=max(num[j]+dp[i][j-1][1],num[j]-max(dp[i][j-1][0],dp[i][j-1][1]));
     }
    printf("%d\n",max(dp[1]
[0],dp[1]
[1]));
  }
}