HDU 4597 Play Game

最新推荐文章于 2017-10-11 19:08:00 发布

Sterben_Da

最新推荐文章于 2017-10-11 19:08:00 发布

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Play Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1122 Accepted Submission(s): 645

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

Sample Output

53 
105 

//别人写的 
#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;  //没有数字了 
    if(dp[la][lb][ra][rb]) return dp[la][lb][ra][rb];  //之前的记录 
    if(la<=ra)
    {
    // 我选了一个值A 给对手的时候总值是sum-A 分治  自己对手拿的也就是他
	//本局拿的B sum-B给对手也就是我  因为每个人拿都是最优解 所以同个递归策略 
        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));  //我开始选的时候的状态 
    }

    return 0;
}













/*\ 自己写的  结构对了  思想不对   
#include <iostream>
#include <cstdio>
#include <string.h>
#include <algorithm>
#include <cstdlib>
using namespace std;
int a[30],b[30];
int dp[22][22][22][22];
//int atail,ahead,bhead,btail;
int dfs(int ahead,int atail,int bhead,int btail){
	if(dp[ahead][atail][bhead][btail])
	return dp[ahead][atail][bhead][btail];
	int ans=0;
	if(ahead>atail&&bhead>btail)
	 return 0;
	 if(ahead==atail&&bhead>btail)
	 return a[ahead];
	 if(ahead>atail&&bhead==btail)
	 return b[bhead];
		dp[ahead][atail][bhead][btail]=
		max(dfs(ahead+1,atail,bhead,btail)+a[ahead],
		max(dfs(ahead,atail-1,bhead,btail)+a[atail],
		max(dfs(ahead,atail,bhead+1,btail)+b[bhead],
		dfs(ahead,atail,bhead,btail-1)+b[btail])));

	return dp[ahead][atail][bhead][btail];
}
int main(){
	int t,n,sum,i;
	scanf("%d",&t);
	while(t--){
		sum=0;
		scanf("%d",&n);
		for(i=0;i<n;i++){
			scanf("%d%d",&a[i],&b[i]);
			sum+=a[i]+b[i];
		}		
	//	atail=0,btail=0,ahead=n-1,bhead=n-1;
		memset(dp,0,sizeof(dp));
		printf("%d\n",max(dfs(1,n-1,0,n-1)+a[0],max(dfs(0,n-2,0,n-1)+a[n-1],
		max(dfs(0,n-1,1,n-1)+b[0],dfs(0,n-1,0,n-2)+b[n-1]))));
	}
	return 0;
}
*/