1031 - Easy Game 简单区间DP

1031 - Easy Game

 

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 A starts 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 than 4digits.

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

 算法分析：

题意：

给一个序列，两个人轮流在序列的两边取任意个数的number，但每次只能从选定的那一边取，问取得数字的和的较大者比较小者多多少？

分析：

dp[i][j]表示区间i到j先手最大比后手多多少分，每一段都这么处理，用小段推导大段。

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

阶段：区间长度

状态：左右端点

决策：划分点k的最佳位置

状态转移方程 ：

dp[i][j] = max(dp[i][j], sum[k] - sum[i-1] - dp[k+1][j] ,sum[j] - sum[k] - dp[i][k]);

初值：dp[i][i]=0,  dp[i][j] = sum[j]-sum[i-1]

目标：dp[1][n]

 

代码实现：

#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;
}