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]。
1 #include <stdio.h> 2 #include <cstring> 3 #include <cstdlib> 4 #include <algorithm> 5 #include <cmath> 6 7 using namespace std; 8 #define lson o<<1 9 #define rson o<<1|1 10 #define max(a,b) (a)>(b)?(a):(b) 11 #define min(a,b) (a)<(b)?(a):(b) 12 #define INF 200000000 13 14 typedef long long ll; 15 int dp[60][50100]; 16 int dfs(int one,int tot){ 17 if(dp[one][tot]>-1)return dp[one][tot]; 18 if(tot==1)return dp[one][tot]=dfs(one+1,0); 19 if(one==0)return dp[one][tot]=tot&1; 20 if(tot>0&&!dfs(one,tot-1))return dp[one][tot]=1; 21 if(!dfs(one-1,tot))return dp[one][tot]=1; 22 if(tot&&!dfs(one-1,tot+1))return dp[one][tot]=1; 23 if(one>=2) 24 if((tot&&!dfs(one-2,tot+3))||(!tot&&!dfs(one-2,tot+2))) 25 return dp[one][tot]=1; 26 27 return dp[one][tot]=0; 28 } 29 int main(){ 30 int t,cs,n; 31 scanf("%d",&t); 32 memset(dp,-1,sizeof dp); 33 dp[0][0]=0; 34 for(cs=1;cs<=t;cs++){ 35 scanf("%d",&n); 36 int x,one=0,tot=0; 37 for(int i=0;i<n;i++){ 38 scanf("%d",&x); 39 if(x==1)one++; 40 else 41 tot+=x+1; 42 } 43 if(tot)tot--; 44 printf("Case #%d: ",cs); 45 if(dfs(one,tot))printf("Alice\n"); 46 else printf("Bob\n"); 47 } 48 return 0; 49 }
