Problem - C - Codeforces

思路：搜索，判断合法性。从起始态用搜索进行模拟，这样可以避免后面判断合法性这一繁琐的步骤。用一个map进行映射当前态及对应的结果。剪枝：如果当前字符串已经被搜索过，则直接跳过去。

代码实现：

void solve() {array<string, 3> st;cin>>st[0]>>st[1]>>st[2];array<string, 3> ph;ph[0] = "...";ph[1] = "..."; ph[2] = "...";map<array<string,3>, string> mp;// 判断是否赢auto check = [&](char ch) -> bool {// 行for(int i = 0; i < 3; ++i) {bool ok = true;for(int j = 0; j < 3; ++j) {ok &= ph[i][j] == ch;}if(ok) return 1;}// 列for(int i = 0; i < 3; ++i) {bool ok = true;for(int j = 0; j < 3; ++j) {ok &= ph[j][i] == ch;}if(ok) return 1;}// 对角线if(ph[0][0] == ch && ph[1][1] == ch && ph[2][2] == ch) return 1;if(ph[2][0] == ch && ph[1][1] == ch && ph[0][2] == ch) return 1;return 0;};// 先是A X// B 0// 从起始态进行搜索auto dfs = [&](auto&& self, int step) -> void {if(check('X')) {mp[ph] = "the first player won";return ;} else if(check('0')) {mp[ph] = "the second player won";return ;} else if(step == 9) {mp[ph] = "draw";return ;}mp[ph] = step % 2 == 0 ? "first" : "second";for(int i = 0; i < 3; ++i) {for(int j = 0; j < 3; ++j) {if(ph[i][j] != '.') continue;ph[i][j] = step % 2 == 0 ? 'X' : '0';if(mp.find(ph) == mp.end())self(self, step + 1);ph[i][j] = '.';}}};dfs(dfs, 0);// 如果在合法态中没有找到就是非法的。if(mp.find(st) == mp.end()) cout<<"illegal";else {cout<<mp[st];}
}