Uva 10891 Game of Sum（区间博弈dp）

10891 - Game of Sum

Time limit: 3.000 seconds

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：给定n个数字，A和B可以从这串数字的两端任意选数字，一次只能从一端选取。

并且A B都尽力使自己选择的结果为最大的。如果A先选择，则A B差值最大是多少。

思路：dp[ i ][ j ] 表示先手玩家在区间 [ i , j ] 所能取得的石子数量的最大值，那么
dp[ i ][ j ] = sum[ j ] - sum[ i - 1 ] - min(dp[ i ][ k ] , dp[ k + 1 ][ j ]) ,其中sum为前缀和。

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int inf=1<<30;
const int maxn=110;

int dp[maxn][maxn],a[maxn],sum[maxn],n;

void initial()
{
for(int i=0;i<=n;i++)
for(int j=0;j<=n;j++)
dp[i][j]=-inf;
sum[0]=0;
}

void input()
{
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
}

int DP(int l,int r)
{
if(dp[l][r]!=-inf)  return dp[l][r];
if(l==r)  return dp[l][r]=a[l];
int ans=sum[r]-sum[l-1];
for(int i=l;i<r;i++)
ans=max(ans,sum[r]-sum[l-1]-min(DP(l,i),DP(i+1,r)));
return dp[l][r]=ans;
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n==0)  break;
initial();
input();
DP(1,n);
printf("%d\n",dp[1]
-(sum
-dp[1]
));
}
return 0;
}