lightoj 1031 - Easy Game（区间DP）

You are playing a two player game. Initially there are n integer numbers in an array and player 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 Astarts the game then how much more point can player A get than player B?

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer N (1 ≤ N ≤ 100) denoting the size of the array. The next line contains N space separated integers. You may assume that no number will contain more than4 digits.

Output

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

Sample Input

Output for Sample Input

2

 

4

4 -10 -20 7

 

4

1 2 3 4

Case 1: 7

Case 2: 10

给你一列数字，你可以取最左边连续的几个，或者最右边连续的几个，问你第一个人取得的数字和最大能比第二个大多少。

dp[i][j]表示区间i到j先手最多比后手多多少分，每一段都这么处理，那么dp[1][n]就是第一个人比第二个人多的分数。

就是找到一点k，使得dp[l][k]-dp[k+1][r]或者dp[k+1][r]-dp[l][k]最大。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int main(void){    int T,n,i,j,k,l;    int sum[110];    int dp[110][110];    scanf("%d",&T);    int cas = 1;    while(T--)    {        scanf("%d",&n);        sum[0] = 0;        for(i=1;i<=n;i++)        {            int x;            scanf("%d",&x);            sum[i] = sum[i-1] + x;        }        for(l=1;l<=n;l++)        {            for(i=1;i+l-1<=n;i++)            {                j = i + l - 1;                dp[i][j] = sum[j]-sum[i-1];                for(k=i;k<j;k++)                {                    int t = max(sum[k] - sum[i-1] - dp[k+1][j],sum[j] - sum[k] - dp[i][k]);                    dp[i][j] = max(dp[i][j],t);                }            }        }        printf("Case %d: %d\n",cas++,dp[1][n]);    }    return 0;}