HDU 4597 Play Game （DP，记忆化搜索）

Play Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 24    Accepted Submission(s): 18


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
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
 

 

 

 

 

只有每种情况记忆化搜索下就可以了

 

 

 1 /* ***********************************************
 2 Author        :kuangbin
 3 Created Time  :2013/8/24 12:37:04
 4 File Name     :F:\2013ACM练习\比赛练习\2013通化邀请赛\1008.cpp
 5 ************************************************ */
 6 
 7 #include <stdio.h>
 8 #include <string.h>
 9 #include <iostream>
10 #include <algorithm>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 int a[30],b[30];
21 int sum1[30];
22 int sum2[30];
23 int dp[30][30][30][30];
24 int solve(int l1,int r1,int l2,int r2)
25 {
26     if(dp[l1][r1][l2][r2] != -1)return dp[l1][r1][l2][r2];
27     if(l1 > r1 && l2 > r2)
28         return dp[l1][r1][l2][r2] = 0;
29     int ans = 0;
30     int sum = 0;
31     if(l1 <= r1)
32         sum += sum1[r1] - sum1[l1-1];
33     if(l2 <= r2)
34         sum += sum2[r2] - sum2[l2-1];
35     if(l1 <= r1)
36     {
37         ans = max(ans,sum - solve(l1+1,r1,l2,r2));
38         ans = max(ans,sum - solve(l1,r1-1,l2,r2));
39     }
40     if(l2 <= r2)
41     {
42         ans = max(ans,sum - solve(l1,r1,l2+1,r2));
43         ans = max(ans,sum - solve(l1,r1,l2,r2-1));
44     }
45     return dp[l1][r1][l2][r2] = ans;
46 }
47 int main()
48 {
49     //freopen("in.txt","r",stdin);
50     //freopen("out.txt","w",stdout);
51     int n;
52     int T;
53     scanf("%d",&T);
54     while(T--)
55     {
56         scanf("%d",&n);
57         sum1[0] = sum2[0] = 0;
58         for(int i = 1;i <= n;i++)
59         {
60             scanf("%d",&a[i]);
61             sum1[i] = sum1[i-1] + a[i];
62         }
63         for(int i = 1;i <= n;i++)
64         {
65             scanf("%d",&b[i]);
66             sum2[i] = sum2[i-1] + b[i]; 
67         }
68         memset(dp,-1,sizeof(dp));
69         printf("%d\n",solve(1,n,1,n));
70     }
71     return 0;
72 }

 

 

 

 

posted on 2013-08-24 18:41  kuangbin  阅读(1736)  评论(0)  编辑  收藏  举报

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