poj 2068 Nim（博弈树）

Nim

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 1501		Accepted: 845

Description

Let's play a traditional game Nim. You and I are seated across a table and we have a hundred stones on the table (we know the number of stones exactly). We play in turn and at each turn, you or I can remove on to four stones from the heap. You play first and the one who removed the last stone loses.
In this game, you have a winning strategy. To see this, you first
remove four stones and leave 96 stones. No matter how I play, I will end
up with leaving 92 - 95 stones. Then you will in turn leave 91 stones
for me (verify this is always possible). This way, you can always leave
5k+1 stones for me and finally I get the last stone, sigh. If we
initially had 101 stones, on the other hand, I have a winning strategy
and you are doomed to lose.

Let's generalize the game a little bit. First, let's make it a team
game. Each team has n players and the 2n players are seated around the
table, with each player having opponents at both sides. Turn around the
table so the two teams play alternately. Second, let's vary the maximum
number of stones each player can take. That is, each player has his/her
own maximum number of stones he/she can take at each turn (The minimum
is always one). So the game is asymmetric and may even be unfair.

In general, when played between two teams of experts, the outcome of
a game is completely determined by the initial number of stones and the
maximum number of stones each player can take at each turn. In other
words, either team has a winning strategy.

You are the head-coach of a team. In each game, the umpire shows
both teams the initial number of stones and the maximum number of stones
each player can take at each turn. Your team plays first. Your job is,
given those numbers, to instantaneously judge whether your team has a
winning strategy.

Incidentally, there is a rumor that Captain Future and her officers
of Hakodate-maru love this game, and they are killing their time playing
it during their missions. You wonder where the stones are? Well, they
do not have stones but do have plenty of balls in the fuel containers!

Input

The
input is a sequence of lines, followed by the last line containing a
zero. Each line except the last is a sequence of integers and has the
following format.

n S M1 M2 . . . M2n

where n is the number of players in a team, S the initial number of
stones, and Mi the maximum number of stones ith player can take. 1st,
3rd, 5th, ... players are your team's players and 2nd, 4th, 6th, ... the
opponents. Numbers are separated by a single space character. You may
assume 1 <= n <= 10, 1 <= Mi <= 16, and 1 <= S < 2^13.

Output

The output should consist of lines each containing either a one, meaning your team has a winning strategy, or a zero otherwise.

Sample Input

1 101 4 4
1 100 4 4
3 97 8 7 6 5 4 3
0

Sample Output

0
1
1

Source

Japan 2001

【思路】

博弈

构造博弈树，一个局面必胜当且仅当后继局面有至少一个必败局面，一个局面必败当且仅当后继局面都为必胜局面。

记忆化搜索即可。

【代码】

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define FOR(a,b,c) for(int a=(b);a<=(c);a++)
 using namespace std;

 const int N =  1e4;

 int f[][N],a[];
 int n,m;

 int dfs(int r,int tot) {
     if(r==*n+) r=;
     int &ans=f[r][tot];
     if(ans!=-) return ans;
     if(tot==) return ans=;
     if(tot<=a[r]) return ans=;
     FOR(i,,a[r])
         if(!dfs(r+,tot-i)) return ans=;
     return ans=;
 }

 int main() {
     while(scanf("%d",&n)== && n) {
         scanf("%d",&m);
         FOR(i,,*n) scanf("%d",&a[i]);
         memset(f,-,sizeof(f));
         if(dfs(,m)) puts("");
         else puts("");
     }
     return ;
 }