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  • hdu 4597 Play Game(区间dp,记忆化搜索)

    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
     
     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
    ",dfs(1,n,1,n));
    72     }
    73     return 0;
    74 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/5000726.html
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