hdu4597Play Game(记忆化）

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通化邀请赛的题 貌似不怎么难

记忆化DP 状态方程类似于2维的 只是变成了4维  每次有四种方向去搜 取最棒的

判断好边界条件

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 using namespace std;

 7 int p1[25],p2[25],sum1[25],sum2[25];

 8 int dp[25][25][25][25];

 9 int dfs(int i1,int j1,int i2,int j2)

10 {

11     if(dp[i1][j1][i2][j2]!=-1)

12     return dp[i1][j1][i2][j2];

13     if(i1>j1&&i2>j2)

14     return dp[i1][j1][i2][j2] = 0;

15     int d1=0,d2=0,d3=0,d4=0;

16     int s1 = sum1[j1]-sum1[i1-1];

17     int s2 = sum2[j2]-sum2[i2-1];

18     if(i1<=j1)

19     {

20         d1 = s2+sum1[j1-1]-sum1[i1-1]-dfs(i1,j1-1,i2,j2)+p1[j1];

21         d2 = s2+sum1[j1]-sum1[i1]-dfs(i1+1,j1,i2,j2)+p1[i1];

22     }

23     if(i2<=j2)

24     {

25         d3 = s1+sum2[j2-1]-sum2[i2-1]-dfs(i1,j1,i2,j2-1)+p2[j2];

26         d4 = s1+sum2[j2]-sum2[i2]-dfs(i1,j1,i2+1,j2)+p2[i2];

27     }

28     dp[i1][j1][i2][j2] = max(max(d1,d2),max(d3,d4));

29     return dp[i1][j1][i2][j2];

30 }

31 int main()

32 {

33     int i,n,t;

34     scanf("%d",&t);

35     while(t--)

36     {

37         cin>>n;

38         memset(sum1,0,sizeof(sum1));

39         memset(sum2,0,sizeof(sum2));

40         memset(dp,-1,sizeof(dp));

41         for(i = 1 ; i <= n ;i++)

42         {

43             scanf("%d",&p1[i]);

44             sum1[i] = sum1[i-1]+p1[i];

45         }

46         for(i = 1 ; i<= n ; i++)

47         {

48             scanf("%d",&p2[i]);

49             sum2[i] = sum2[i-1]+p2[i];

50         }

51         int ans = dfs(1,n,1,n);

52         cout<<ans<<endl;

53     }

54     return 0;

55 }

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