hdu4597Play Game(记忆化）
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 }
