HDU 4155 The Game of 31 （博弈）

题意:

1,2,3,4,5,6各有4张一共24张牌,2个人拿牌,要求牌的总和不能超过31,超过了不能拿,谁不能拿谁输

分析:

博弈基本思想:

当前是必胜态,当且仅当有一个后继是必败态

当前是必败态,当且仅当所有后继是必胜态

没有后继的是必败态

其实第一个拿到31的是必胜态,写的时候反过来看起来跑得快

代码:

//
//  Created by TaoSama on 2015-11-21
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int card[10];
char s[20];

bool dfs(int sum) {
if(sum == 31) return false;
for(int i = 1; i <= 6; ++i) {
if(sum + i > 31) break;
if(!card[i]) continue;
--card[i];
bool F = dfs(sum + i);
++card[i];
if(!F) return true;
}
return false;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%s", s) == 1) {
for(int i = 1; i <= 6; ++i) card[i] = 4;
int sum = 0;
int n = strlen(s);
for(int i = 0; i < n; ++i) {
--card[s[i] - '0'];
sum += s[i] - '0';
}
printf("%s ", s);
//这里必败和必胜反过来了
if(dfs(sum)) printf("%c\n", "AB"[n & 1]); //必败
else printf("%c\n", "BA"[n & 1]);  //必胜
}
return 0;
}