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HDU4597：Play Game(记忆化)

2024-07-14 15:01:34   225人阅读

Problem Description
Alice and Bob are playing a game. There are two piles of cards. There are N cards in each pile, and each card has a score. They take turns to pick up the top or bottom card from either pile, and the score of the card will be added to his total score. Alice and Bob are both clever enough, and will pick up cards to get as many scores as possible. Do you know how many scores can Alice get if he picks up first?
 

Input
The first line contains an integer T (T≤100), indicating the number of cases. 
Each case contains 3 lines. The first line is the N (N≤20). The second line contains N integer ai (1≤ai≤10000). The third line contains N integer bi (1≤bi≤10000).
 

Output
For each case, output an integer, indicating the most score Alice can get.
 

Sample Input
2 
 
1 
23 
53 
 
3 
10 100 20 
2 4 3
 

Sample Output
53 
105 
dp[la][ra][lb][rb]记录的是在a的区间只剩下la~ra，b的区间只剩下lb~rb的时候，Alice能得到的最大值
那么只需要考虑四种不同的取法并从中取得最优的方案，sum-(Alice上一个状态中Bob拿的值)中取最大
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int n,dp[23][23][23][23],a[23],b[23];

int dfs(int la,int ra,int lb,int rb,int sum)
{
    int maxn = 0;
    if(la>ra && lb>rb) return 0;
    if(dp[la][lb][ra][rb]) return dp[la][lb][ra][rb];
    if(la<=ra)
    {
        maxn = max(maxn,sum-dfs(la+1,ra,lb,rb,sum-a[la]));
        maxn = max(maxn,sum-dfs(la,ra-1,lb,rb,sum-a[ra]));
    }
    if(lb<=rb)
    {
        maxn = max(maxn,sum-dfs(la,ra,lb+1,rb,sum-b[lb]));
        maxn = max(maxn,sum-dfs(la,ra,lb,rb-1,sum-b[rb]));
    }
    dp[la][lb][ra][rb] = maxn;
    return maxn;
}

int main()
{
    int t,i,j,sum;
    scanf("%d",&t);
    while(t--)
    {
        sum = 0;
        scanf("%d",&n);
        for(i = 1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        for(i = 1;i<=n;i++)
        {
            scanf("%d",&b[i]);
            sum+=b[i];
        }
        memset(dp,0,sizeof(dp));
        printf("%d\n",dfs(1,n,1,n,sum));
    }

    return 0;
}



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