UVA

题目连接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19461

　　Game of sum

Description

This is a two player game. Initially there are n integer numbers in an array and players A and B getchance to take them alternatively. Each player can take one or more numbers from the left or right endof the array but cannot take from both ends at a time. He can take as many consecutive numbers ashe 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. Eachplayer 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

 4

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

Sample Output

7

10

Hint

题意：

 　　给你一个长度n的数组，AB两人轮流开始选一段数，至少一个，但是只能从两端开始，问你最后每个人选的权值和A最多比B多多少，A,B一样聪明

题解：

　　区间DP，dp[i][j]表示从i，j内最佳答案

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2e2+10,inf = 2e9, mod = 1e9+7;
typedef long long ll;

const ll INF = 1e18;

ll dp[N][N],a[N],sum[N],n;

ll dfs(int l,int r)
{
    if(dp[l][r]!=INF)
    {
        return dp[l][r];
    }
    ll &ret=dp[l][r];
    ret=-INF;
    for(int i=l;i<r;i++)
    {
        ret = max(ret,sum[i]-sum[l-1] - dfs(i+1,r));
    }
    for(int i=r;i>l;i--)
    {
        ret = max(ret,sum[r]-sum[i-1] - dfs(l,i-1));
    }
    ret = max(ret,sum[r]-sum[l-1]);
    return ret;
}
int main()
{
    while(scanf("%lld",&n)!=EOF)
    {
        if(n==0)break;
        for(int i=0;i<=n+100;i++)
            for(int j=0;j<=n+100;j++) dp[i][j]=INF;
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
            sum[i]=(sum[i-1]+a[i]);
        }
        printf("%lld
",dfs(1,n));
    }
    return 0;
}