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最新推荐文章于 2024-08-04 16:36:27 发布

转载最新推荐文章于 2024-08-04 16:36:27 发布 · 108 阅读

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原文链接：http://www.cnblogs.com/Staceyacm/p/11248746.html

本文探讨了一个二人轮流从数组两端取数的游戏策略。通过动态规划算法，计算了在双方都采用最优策略的情况下，先手玩家相较于后手玩家最多能获得多少额外的分数。文章提供了详细的算法实现代码，并通过样例输入输出展示了算法的有效性。

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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 Nspace separated integers. You may assume that no number will contain more than 4 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

4 -10 -20 7

1 2 3 4

Sample Output

Case 1: 7

Case 2: 10

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#define Inf 0x3f3f3f3f
const int maxn=1e5+5;
typedef long long ll;
using namespace std;

int dp[205][205];
int sum[205];
int n;
int dfs(int l,int r)
{
    if(dp[l][r]!=-Inf)
    {
        return dp[l][r];
    }
    if(l>r)
    {
        return 0;
    }
    int ans=-Inf;
    for(int t=1;t<=n;t++)
    {
        if(l+t<=r+1)
        ans=max(ans,sum[l+t-1]-sum[l-1]-dfs(l+t,r));
    }
    for(int t=1;t<=n;t++)
    {
        if(r-t>=l-1)
        ans=max(ans,sum[r]-sum[r-t]-dfs(l,r-t));    
    }
//    cout<<l<<" "<<r<<" "<<ans<<endl;
    return dp[l][r]=ans;
}
int main()
{
   int T;
   cin>>T;
   int cnt=1;
   while(T--)
   {
       scanf("%d",&n);
       for(int t=0;t<=n;t++)
       {
           for(int j=0;j<=n;j++)
           {
               dp[t][j]=-Inf;
        }
    }
       memset(sum,0,sizeof(sum));
       int x;
       for(int t=1;t<=n;t++)
       {
           scanf("%d",&x);
           dp[t][t]=x;
           sum[t]=sum[t-1]+x;
    }
    dfs(1,n);
    printf("Case %d: %d\n",cnt++,dp[1][n]);
   }
   return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/11248746.html