HDU 4597（记忆化搜索 dfs 参考）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4597

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 a _i (1≤a _i≤10000). The third line contains N integer b _i (1≤b _i≤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

 

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

PS:

记忆化搜索。

将两个人抽象出来可以看做是一样的，因为他们都具备共同的性质：都很聪明（即在当前状态下保证是自己拿的最多）。

这样就好处理多了，假设dfs()能生成一个在当前状态下的最优解，用数组dp记录下来，然后递归完成子问题就形成记忆化搜索了。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 32;
int a[maxn], b[maxn], suma[maxn], sumb[maxn];
int dp[maxn][maxn][maxn][maxn];
//dp[x][y][i][j]表示当前玩家从a堆的x~y,b堆的i~j能获得的最大价值 
int dfs(int x, int y, int i, int j)
{
    if(dp[x][y][i][j])
    {
        return dp[x][y][i][j];
    }
    if(x > y && i > j)
    {
        return 0;
    }
    int maxa = 0, maxb = 0;
    if(x <= y)
    {
        maxa = max(a[x]+dfs(x+1,y,i,j),a[y]+dfs(x,y-1,i,j));
    }
    if(i <= j)
    {
        maxb = max(b[i]+dfs(x,y,i+1,j),b[j]+dfs(x,y,i,j-1));
    }
    dp[x][y][i][j] = suma[y]-suma[x-1]+sumb[j]-sumb[i-1]-max(maxa,maxb);
    
    //区间和减去对手所取的剩下的就为当前玩家的
    return dp[x][y][i][j];
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        memset(suma, 0,sizeof(suma));
        memset(sumb, 0,sizeof(sumb));
        int n;
        scanf("%d",&n);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&a[i]);
            suma[i] = suma[i-1]+a[i];
        }
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&b[i]);
            sumb[i] = sumb[i-1]+b[i];
        }
        int ans = suma[n]+sumb[n]-dfs(1,n,1,n);
        printf("%d\n",ans);
    }
    return 0;
}