hdu4597 Play Game

最新推荐文章于 2020-08-27 12:21:44 发布

weixin_30613343

最新推荐文章于 2020-08-27 12:21:44 发布

阅读量56

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/pangblog/p/3281258.html

本文探讨了一个双堆卡牌游戏的策略问题，玩家轮流抽取卡牌以最大化得分。通过区间DP算法，作者详细解释了如何计算最优得分，并提供了代码实现。此问题涉及动态规划和区间操作，适合寻求解决复杂决策问题的读者。

Play Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 54 Accepted Submission(s): 36

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

Recommend

liuyiding

这题就是一个区间dp,因为，每一次取都，可以取两个队列的头和尾，所以，就是一个二维的区间dp,我们用dp[al][ar][bl][br]表示，从第一个数列从al 到ar,第二个数列从，bl到br,当前这个人具有第一选择权的最大分值，那么我们，可以取两个队列的头和尾，就有4个状态转移方程了！用一个记忆化搜索就可以了！

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define MAXN 26
int suma[MAXN],sumb[MAXN],pa[MAXN],pb[MAXN],dp[MAXN][MAXN][MAXN][MAXN];
int dfs(int al,int ar,int bl ,int br)
{
    if(dp[al][ar][bl][br]!=-1)
    return dp[al][ar][bl][br];
    dp[al][ar][bl][br]=0;
    if(al<=ar)
    dp[al][ar][bl][br]=suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al+1,ar,bl,br);
    if(al<=ar)
    dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar-1,bl,br));
    if(bl<=br)
    dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar,bl+1,br));
    if(bl<=br)
    dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar,bl,br-1));
    return dp[al][ar][bl][br];
}
int main ()
{
    int n,i,tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d",&n);
        suma[0]=sumb[0]=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&pa[i]);
            suma[i]=suma[i-1]+pa[i];
        }
        for(i=1;i<=n;i++)
        {
            scanf("%d",&pb[i]);
            sumb[i]=sumb[i-1]+pb[i];
        }
        memset(dp,-1,sizeof(dp));
        printf("%d\n",dfs(1,n,1,n));
    }
    return 0;
}

转载于:https://www.cnblogs.com/pangblog/p/3281258.html