https://leetcode.com/problems/can-i-win/description/
参考了这里 https://leetcode.com/problems/can-i-win/discuss/95320/7-liner-C++-beat-98.4-DFS-with-early-termination-check-(detailed-explanation)
看数据规模能猜测到是状压。但是博弈思维还差点。
如果下一个状态有一个是必败的,那么当前状态就是必胜的。dp[i]表示当前状态未搜索/必胜/必败
int dp[1<<20+1] = {}; class Solution { public: bool canIWin(int maxv, int target) { if ((maxv+1) * maxv / 2 < target) return false; memset(dp, 0, sizeof(dp)); return target < 2 ? true : dfs(maxv, target, 0); } bool dfs(int maxv, int target, int state) { if (target <= 0) return false; if (dp[state] != 0) return dp[state] == 1; for (int i = 0; i < maxv; i++) { if( !(state & 1<<i) && !dfs(maxv, target - (i+1) , state | 1<<i) ) { //找到了一种必输的路径,让下一个状态必输 return dp[state] = 1; } } //所有的下一个状态都是必胜结局,那么现在这个肯定是必输的 return !(dp[state ] = -1); } };