hdu4597 （博弈dp）play game

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 
思路：dp[al][ar][bl][br]表示在pile1的数还剩下从al到ar（开区间），pile2的数还剩下bl到br的情况下,先手取得的最大值。
所以，每次选择当前所有减去下一个人的最大值（四个方向），即可
理解：你的最大等于剩余所有减去下一个人取最大（奇偶奇偶奇。。。。。）

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;const int maxn = 20 + 5;int a[maxn];int b[maxn];int suma[maxn],sumb[maxn];int dp[maxn][maxn][maxn][maxn];int dfs(int al,int ar,int bl,int br){    if(dp[al][ar][bl][br] != -1) return dp[al][ar][bl][br];    dp[al][ar][bl][br] = 0;    if(al < ar-1) dp[al][ar][bl][br] = a[al+1] + (suma[ar-1]-suma[al+1]+sumb[br-1]-sumb[bl]-dfs(al+1,ar,bl,br));    if(al < ar-1) dp[al][ar][bl][br] = max(dp[al][ar][bl][br],a[ar-1] + (suma[ar-2]-suma[al]+sumb[br-1]-sumb[bl]-dfs(al,ar-1,bl,br)));    if(bl < br-1) dp[al][ar][bl][br] = max(dp[al][ar][bl][br],b[bl+1] + (suma[ar-1]-suma[al]+sumb[br-1]-sumb[bl+1]-dfs(al,ar,bl+1,br)));    if(bl < br-1) dp[al][ar][bl][br] = max(dp[al][ar][bl][br],b[br-1] + (suma[ar-1]-suma[al]+sumb[br-2]-sumb[bl]-dfs(al,ar,bl,br-1)));    return  dp[al][ar][bl][br];}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        suma[0] = 0;sumb[0] = 0;        for(int i = 1;i <= n;i++)        {            scanf("%d",&a[i]);            suma[i] = suma[i-1] + a[i];        }        for(int i = 1;i <= n;i++)        {            scanf("%d",&b[i]);            sumb[i] = sumb[i-1]+b[i];        }        memset(dp,-1,sizeof(dp));        printf("%d\n",dfs(0,n+1,0,n+1));    }    return 0;}