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[60][50100];
int dfs(int one,int tot){
if(dp[one][tot]>-1)return dp[one][tot];
if(tot==1)return dp[one][tot]=dfs(one+1,0);
if(one==0)return dp[one][tot]=tot&1;
if(tot>0&&!dfs(one,tot-1))return dp[one][tot]=1;
if(!dfs(one-1,tot))return dp[one][tot]=1;
if(tot&&!dfs(one-1,tot+1))return dp[one][tot]=1;
if(one>=2)
if((tot&&!dfs(one-2,tot+3))||(!tot&&!dfs(one-2,tot+2)))
return dp[one][tot]=1;

return dp[one][tot]=0;
}
int main(){
int t,cs,n;
scanf("%d",&t);
memset(dp,-1,sizeof dp);
dp[0][0]=0;
for(cs=1;cs<=t;cs++){
scanf("%d",&n);
int x,one=0,tot=0;
for(int i=0;i<n;i++){
scanf("%d",&x);
if(x==1)one++;
else
tot+=x+1;
}
if(tot)tot--;
printf("Case #%d: ",cs);
if(dfs(one,tot))printf("Alice\n");
else printf("Bob\n");
}
return 0;
}