POJ 2311 Cutting Game (博弈)

题意：给定一个长方形纸张，每次只能水平或者垂直切，如果切到1*1的方格就胜，问先手胜还是负。

析：根据Nim游戏可知，我们可以分别求出每个子游戏的和，就是答案，所以我们就枚举每一种切法，然后求出SG函数，那么就能得到答案。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 200 + 10;
const int mod = 1e6 + 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}
int dp[maxn][maxn];

int dfs(int r, int c){
  int &ans = dp[r][c];
  if(ans >= 0)  return ans;
  set<int> sets;
  for(int i = 2; r - i > 1; ++i) sets.insert(dfs(r-i, c) ^ dfs(i, c));
  for(int i = 2; c - i > 1; ++i) sets.insert(dfs(r, i) ^ dfs(r, c-i));
  int res = 0;
  while(sets.count(res))  ++res;
  return ans = res;
}

int main(){
  memset(dp, -1, sizeof dp);
  while(scanf("%d %d", &m, &n) == 2){
    printf("%s
", dfs(m, n) ? "WIN" : "LOSE");
  }
  return 0;
}