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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 54 Accepted Submission(s): 36
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 a i (1≤a i≤10000). The third line contains N integer b i (1≤b i≤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
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liuyiding
这题就是一个区间dp,因为，每一次取都，可以取两个队列的头和尾，所以，就是一个二维的区间dp,我们用dp[al][ar][bl][br]表示，从第一个数列从al 到ar,第二个数列从，bl到br,当前这个人具有第一选择权的最大分值，那么我们，可以取两个队列的头和尾，就有4个状态转移方程了！用一个记忆化搜索就可以了！
#include#include #include #include using namespace std;#define MAXN 26int suma[MAXN],sumb[MAXN],pa[MAXN],pb[MAXN],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) dp[al][ar][bl][br]=suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al+1,ar,bl,br); if(al<=ar) dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar-1,bl,br)); if(bl<=br) dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar,bl+1,br)); if(bl<=br) dp[al][ar][bl][br]=max(dp[al][ar][bl][br],suma[ar]-suma[al-1]+sumb[br]-sumb[bl-1]-dfs(al,ar,bl,br-1)); return dp[al][ar][bl][br];}int main (){ int n,i,tcase; scanf("%d",&tcase); while(tcase--) { scanf("%d",&n); suma[0]=sumb[0]=0; for(i=1;i<=n;i++) { scanf("%d",&pa[i]); suma[i]=suma[i-1]+pa[i]; } for(i=1;i<=n;i++) { scanf("%d",&pb[i]); sumb[i]=sumb[i-1]+pb[i]; } memset(dp,-1,sizeof(dp)); printf("%d\n",dfs(1,n,1,n)); } return 0;}