本文共 2081 字,大约阅读时间需要 6 分钟。
给出n个字符串,进行k次游戏,每次游戏输家下次作为先手,游戏规则为每次放一个字母,导致当前构造的字符串是给定的任意一个字符串的前缀,不能操作时为输,赢得第k次比赛的人会取得最终的胜利,问两人都采取最优策略的情况下,谁会赢得比赛。
#include#include #include #include #define MAX 100007using namespace std;int n,k,cc;char s[MAX];struct Node{ int branch[26]; Node ( ) { memset ( branch , -1 , sizeof ( branch)); } int win,lose;}p[MAX<<1];void add ( ){ int u = 0; int m = strlen ( s ); for ( int i = 0 ; i < m ; i++ ) { int x = s[i]-'a'; if ( p[u].branch[x] == -1 ) p[u].branch[x] = cc++; u = p[u].branch[x]; }}void dfs ( int u , int d ){ int temp1,temp2; temp1 = temp2 = (d&1)?0:1; bool flag = false; for ( int i = 0 ; i < 26 ; i++ ) { int v = p[u].branch[i]; if ( v == -1 ) continue; flag = true; dfs ( v , d+1 ); if ( d&1 ) { temp1 = temp1||p[v].win; temp2 = temp2||p[v].lose; } else { temp1 = temp1&&p[v].win; temp2 = temp2&&p[v].lose; } } if ( !flag ) { p[u].win = p[u].lose = 0; if ( d&1 ) p[u].lose = 1; else p[u].win = 1; } else { p[u].win = temp1; p[u].lose = temp2; }}int main ( ){ while ( ~scanf ( "%d%d" , &n , &k )) { cc = 1; for ( int i = 0 ; i < n ; i++ ) { scanf ( "%s" , s ); add ( ); } dfs ( 0 , 1 ); int win = p[0].win; int lose = p[0].lose; if ( !win ) puts ( "Second"); else if ( lose ) puts ("First"); else { if ( k&1 ) puts ("First"); else puts ("Second"); } }}
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