uvalive 5760 Alice and Bob （组合游戏，dp)

题目链接： http://vjudge.net/problem/viewProblem.action?id=25636

对于>1的堆，必然会被其中一人全部合并。

然后就是二维dp，dp[非1堆的操作数][1堆个数]。

 #include <stdio.h>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #include <cmath>
 
 using namespace std;
 #define lson o<<1
 #define rson o<<1|1
 #define max(a,b) (a)>(b)?(a):(b)
 #define min(a,b) (a)<(b)?(a):(b)
 #define INF 200000000
 
 typedef long long ll;
 int dp[][];
 int dfs(int one,int tot){
     if(dp[one][tot]>-)return dp[one][tot];
     if(tot==)return dp[one][tot]=dfs(one+,);
     if(one==)return dp[one][tot]=tot&;
     if(tot>&&!dfs(one,tot-))return dp[one][tot]=;
     if(!dfs(one-,tot))return dp[one][tot]=;
     if(tot&&!dfs(one-,tot+))return dp[one][tot]=;
     if(one>=)
        if((tot&&!dfs(one-,tot+))||(!tot&&!dfs(one-,tot+)))
            return dp[one][tot]=;
 
     return dp[one][tot]=;
 }
 int main(){
    int t,cs,n;
    scanf("%d",&t);
    memset(dp,-,sizeof dp);
    dp[][]=;
    for(cs=;cs<=t;cs++){
       scanf("%d",&n);
       int x,one=,tot=;
       for(int i=;i<n;i++){
          scanf("%d",&x);
          if(x==)one++;
          else
             tot+=x+;
       }
       if(tot)tot--;
       printf("Case #%d: ",cs);
       if(dfs(one,tot))printf("Alice\n");
       else printf("Bob\n");
    }
    return ;
 }