10891 - Game of Sum

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

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Problem setter: Syed Monowar Hossain

Special Thanks: Derek Kisman, Mohammad Sajjad Hossain

 整数的总和是一定的，不管怎么取，任意时刻的状态都是原始序列的字串，因此我们想到用d[i][j]表示原始序列i->j个元素组成的子序列，在双方都取得最优策越的情况下先手得到的最高分，状态转移时，我们需要枚举从左边取和从右边取，这等价于给对方剩下怎样的序列。则有 d[i][j]=sum(i,j)-min(d[i+1][j],……d[j][j],d[i][j-1],……d[i][i],0),  则ans=d[1][n]-(sum(1,n)-d[1][n])

#include"iostream"
#include"cstring"
#include"cstdio"
#include"algorithm"
using namespace std;
const int ms=1005;
int sum[ms];
int d[ms][ms];
int vis[ms][ms];
int a[ms];
int n;
int dp(int i,int j)
{
    if(vis[i][j])
        return d[i][j];
    int m=0;   //全部取
    vis[i][j]=1;
    for(int k=i+1;k<=j;k++)
        m=min(m,dp(k,j));
    for(int k=i;k<j;k++)
        m=min(m,dp(i,k));
    d[i][j]=sum[j]-sum[i-1]-m;
    return d[i][j]; 
} 
int main()
{
    while(scanf("%d",&n)&&n)
    {
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        memset(vis,0,sizeof(vis));
        printf("%d\n",2*dp(1,n)-sum[n]);
    }
    return 0;
}