HDU-4597 Play Game （区间DP）

Play Game

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

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

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 
   
   
   
   

题目大意：给定两堆均含有n张牌的牌堆，牌上的数字分别是a[i],b[i]，每次都已从任意一堆的顶部或底部拿一张，每个人都用最优策略，求先手能拿到的最大数字和？

设dp[i][j][k][l]表示先手从a数组区间[i,j]中以及b数组区间[k,l]中能选得的最大数字和

则dp[i][j][k][l]可由四个状态转移而来：

①先手拿a[i]，由dp[i+1][j][k][l]转移而来，即 a[i]+sum(i+1,j,k,l)-dp[i+1][j][k][l];

②先手拿a[j]，由dp[i][j-1][k][l]转移而来，即 a[j]+sum(i,j-1,k,l)-dp[i][j-1][k][l];

③先手拿b[k]，由dp[i][j][k+1][l]转移而来，即 b[k]+sum(i,j,k+1,l)-dp[i][j][k+1][l];

④先手拿b[l]，由dp[i][j][k][l-1]转移而来，即 b[l]+sum(i,j,k,l-1)-dp[i][j][k][l-1];

则dp[i][j][k][l]为上述最大值

感觉和只有一堆牌的情况差不多，但是连状态如何转移都知道了，却不知道如何实现，因为存在一堆没有选，而另一堆已选了的状态（这些状态没法表示，不好控制边界），看了题解才知道要用dfs控制边界从而进行状态的转移

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN=10005;

int par[MAXN],ps,pe,n,m,a[25],b[25];
int dp[25][25][25][25];//dp[i][j][k][l]表示先手从a数组区间[i,j]中以及b数组区间[k,l]中能选得的最大数字和

int dfs(int i,int j,int k,int l) {
    if(i>j&&k>l) {
        return 0;
    }
    if(dp[i][j][k][l]!=0) {
        return dp[i][j][k][l];
    }

    int mx=0,sum=0;
    if(i<=j) {
        sum+=a[j]-a[i-1];
    }
    if(k<=l) {
        sum+=b[l]-b[k-1];
    }
    if(i<=j) {
        mx=max(mx,sum-dfs(i+1,j,k,l));
        mx=max(mx,sum-dfs(i,j-1,k,l));
    }
    if(k<=l) {
        mx=max(mx,sum-dfs(i,j,k+1,l));
        mx=max(mx,sum-dfs(i,j,k,l-1));
    }
    return dp[i][j][k][l]=mx;
}

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