HDU 4597 Play Game（DFS，区间DP）

Play Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)

Total Submission(s): 1274 Accepted Submission(s): 737

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[x1][y1][x2][y2]表示当前先手的这个人在x1到y1,x2到y2,两堆的区间，可以获得最大值，dp[x1][y1][x2][y2]的值等于当前的和 sum-(下一个状态，即对方先手可以获得的最大值)。

关于区间DP,可以参照这个博客

http://blog.csdn.net/dacc123/article/details/50885903

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int dp[30][30][30][30];
int a[30];
int b[30];
int n;
int sum;
int dfs(int x1,int y1,int x2,int y2,int sum)
{
    if(dp[x1][y1][x2][y2])
        return dp[x1][y1][x2][y2];
    if((x1>y1)&&(x2>y2))
        return 0;
    int num=0;
    if(x1<=y1)
    {
        num=max(num,sum-dfs(x1+1,y1,x2,y2,sum-a[x1]));
        num=max(num,sum-dfs(x1,y1-1,x2,y2,sum-a[y1]));
    }
    if(x2<=y2)
    {
        num=max(num,sum-dfs(x1,y1,x2+1,y2,sum-b[x2]));
        num=max(num,sum-dfs(x1,y1,x2,y2-1,sum-b[y2]));
    }
    dp[x1][y1][x2][y2]=num;
    return dp[x1][y1][x2][y2];
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        for(int 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;
}