hdu 4597 Play Game(区间dp,记忆化搜索)

最新推荐文章于 2021-10-19 09:31:19 发布

转载最新推荐文章于 2021-10-19 09:31:19 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/UniqueColor/p/5000726.html

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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

Sample Output

53 
105

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 #define PI acos(-1.0)
17 #define ll long long
18 #define eps 1e-10
19 #define MOD 1000000007
20 #define N 26
21 #define inf 1e12
22 int sum1[26],sum2[26];
23 int a[26];
24 int b[26];
25 int dp[26][26][26][26];
26 int dfs(int l1,int r1,int l2,int r2)
27 {
28     if(dp[l1][r1][l2][r2]!=-1)
29        return dp[l1][r1][l2][r2];
30     //if(l1>r1 && l2>r2) return dp[l1][r1][l2][r2]=0;
31     int sum=0;
32     int ans=0;
33     if(l1<=r1)
34     {
35         sum+=sum1[r1]-sum1[l1-1];
36     }
37     if(l2<=r2)
38     {
39         sum+=sum2[r2]-sum2[l2-1];
40     }
41     if(l1<=r1)
42     {
43         ans=max(ans,sum-min(dfs(l1+1,r1,l2,r2),dfs(l1,r1-1,l2,r2)));
44     }
45     if(l2<=r2)
46     {
47         ans=max(ans,sum-min(dfs(l1,r1,l2+1,r2),dfs(l1,r1,l2,r2-1)));
48     }
49     return dp[l1][r1][l2][r2]=ans;
50 }
51 int main()
52 {
53     int t;
54     int n;
55     cin>>t;
56     while(t--)
57     {
58         cin>>n;
59         sum1[0]=sum2[0]=0;
60         for(int i=1;i<=n;i++)
61         {
62             scanf("%d",&a[i]);
63             sum1[i]=sum1[i-1]+a[i];
64         }
65         for(int i=1;i<=n;i++)
66         {
67             scanf("%d",&b[i]);
68             sum2[i]=sum2[i-1]+b[i];
69         }
70         memset(dp,-1,sizeof(dp));
71         printf("%d\n",dfs(1,n,1,n));
72     }
73     return 0;
74 }

View Code

转载于:https://www.cnblogs.com/UniqueColor/p/5000726.html