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 << endlconst 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;}