HDU 4597 Play Game（记忆化搜索）

最新推荐文章于 2020-10-20 20:23:46 发布

原创最新推荐文章于 2020-10-20 20:23:46 发布 · 314 阅读

0 ·

CC 4.0 BY-SA版权

DFS算法专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个涉及两个玩家的游戏策略问题，玩家轮流从两堆卡片中取卡以获得最高分数。文章通过记忆化搜索和动态规划的方法，给出了最优策略下先手玩家能获得的最大分数。

Play Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1704 Accepted Submission(s): 1014

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

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

记忆化搜索，dp[l][r][l1][r1]表示第一、第二堆的边界内 b可以取到的最大值，因为b是最后取完的点，比较容易判断结束，maxa，maxb表示 a b在当前的范围内取得值

#include <stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
using namespace std;
int dp[50][50][50][50];
int a[101];
int b[101];
int suma[1001],sumb[1001];
int dfs(int l,int r,int l1,int r1)
{
	if(dp[l][r][l1][r1]!=-1)
	{
		return dp[l][r][l1][r1];
	}	
	int maxa=0,maxb=0;
	if(l<=r)
	{
		maxa=max(dfs(l+1,r,l1,r1)+a[l],dfs(l,r-1,l1,r1)+a[r]);	
	}
	
	if(l1<=r1)
	{
		maxb=max(dfs(l,r,l1+1,r1)+b[l1],dfs(l,r,l1,r1-1)+b[r1]);	
	}
	dp[l][r][l1][r1]=suma[r]-suma[l-1]+sumb[r1]-sumb[l1-1]-max(maxa,maxb);
	return dp[l][r][l1][r1];
} 
int main(int argc, char *argv[])
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		memset(suma,0,sizeof(suma));
		memset(sumb,0,sizeof(sumb));
		memset(dp,-1,sizeof(dp));	
		int i;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			suma[i]+=suma[i-1]+a[i];
		}
		
		for(i=1;i<=n;i++)
		{
			scanf("%d",&b[i]);
			sumb[i]=sumb[i-1]+b[i];
		}
		dfs(1,n,1,n);
		printf("%d\n",suma[n]+sumb[n]-dp[1][n][1][n]);
		
	}
	return 0;
}