Uva 10891 - Game of Sum dp博弈

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本文介绍了一个两人轮流取数的游戏问题，通过动态规划算法实现最优策略以计算首位玩家相较于第二位玩家能获得的最大得分差。文章提供了一段C++代码实现，并附带了详细的输入输出样例。

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 -10 -20 7

1 2 3 4

10

Problem setter: Syed Monowar Hossain

Special Thanks: Derek Kisman, Mohammad Sajjad Hossain

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f(i,j)=sum(i,j)-min(f(k1,j), f(i,k2), 0 ) ( i<k1<=j i<=k2<j )

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#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int maxn=111;
const int INF=1e9;

int n;
int a[maxn];
int f[maxn][maxn];
bool v[maxn][maxn];
int sum[maxn];

int dp(int l,int r)
{
    int ret=0;
    if (v[l][r]) return f[l][r];
    for (int k=l+1;k<=r;k++) ret=min(ret,dp(k,r));
    for (int k=l;k<=r-1;k++) ret=min(ret,dp(l,k));
    ret=sum[r]-sum[l-1]-ret;
    f[l][r]=ret;
    v[l][r]=true;
    return ret;
}

int main()
{
    while (~scanf("%d",&n))
    {
        if (n==0) break;
        memset(v,0,sizeof(v));
        sum[0]=0;
        for (int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        int ans=dp(1,n);
        printf("%d\n",ans*2-sum[n]);
    }
    return 0;
}