HDU-1079 Calendar Game

Calendar Game

给出一个终止日期和开始日期，两个人每次只能有两种方式移动的合法日期，如果日期不合法，则不能移动：

1. 移动到当前日期的下一天

2. 移动到当前日期的下一月同一天

先到达终止日期的获胜，不允许到达终止日期后的日期

博弈论

简单的一个博弈图分析，分析一下必胜态和必输态即可，日期处理上面比较繁琐，状态用map_sg来表示，0表示必输态，1表示必胜态

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>
#include <functional>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <deque>
#include <stack>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
const ll maxn = 2e5 + 10;
const ll inf = 1e17 + 10;
int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool r_year(int year)
{
    return (year % 400 == 0) || (year % 4 == 00 && year % 100 != 0);
}

struct node
{
    int year, month, day;
    node(){}
    node(int _y, int _m, int _d){year = _y; month = _m; day = _d;}
    bool fit()
    {
        if(month == 13) month = 1, year++;
        if(r_year(year)) m[2] = 29;
        bool flag = true;
        if(day > m[month] || day < 1) flag = false;;
        m[2] = 28;
        return flag;
    }
    
    bool operator < (const node& a) const
    {
        if(a.year != year) return year < a.year;
        if(a.month != month) return month < a.month;
        return day < a.day;
    }
    void operator --(int)
    {
        day--;
        if(day == 0)
        {
            month--;
            if(month == 0)
            {
                year--;
                month = 12;
            }
            if(r_year(year)) m[2] = 29;
            day = m[month];
            m[2] = 28;
        }
    }
};
map<node, int>sg;

int main()
{
    node r(2001, 11, 4), l(1900, 1, 0);
    int cur = 1;
    node pre = r, last(2001, 11, 5);
    r--;
    while(l < r)
    {
        int f = 1;
        f = min(f, sg[pre]);
        node nex = r;
        nex.month += 1;
        if(nex.fit() && nex < last)
            f = min(f, sg[nex]);
        if(f) sg[r] = 0;
        else sg[r] = 1;
        pre = r;
        r--;
    }

    int t;
    scanf("%d", &t);
    while(t--)
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        node now(a, b, c);
        printf("%s\n", sg[now] == 0 ? "NO" : "YES");
    }

    return 0;
}