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Codeforces Beta Round #4 (Div. 2 Only) C. Registration system_html/css_WEB-ITnose

WBOY
WBOYOriginal
2016-06-24 11:53:101001browse

这个题感觉还不错,以前字典树写的是最顺手的,这几次比赛屡屡挂在字典树上也是有阴影了啊~~



题目大意:

给出一些字符串,对每个字符串进行查询,若没出现过返回OK,若出现过就生成新字符串,格式为原字符串+数,数为这个字符串第几次重复出现。



解题思路:

字典树,对于每个字符串的插入次数进行计数。



下面是代码:

#include #include #include #include #include #include #include #include #include #include #include #include #define eps 1e-10#define pi acos(-1.0)#define inf 107374182#define inf64 1152921504606846976#define lc l,m,tr 0 ? (x) : -(x))#define clear1(A, X, SIZE) memset(A, X, sizeof(A[0]) * (min(SIZE,sizeof(A))))#define clearall(A, X) memset(A, X, sizeof(A))#define memcopy1(A , X, SIZE) memcpy(A , X ,sizeof(X[0])*(SIZE))#define memcopyall(A, X) memcpy(A , X ,sizeof(X))#define max( x, y )  ( ((x) > (y)) ? (x) : (y) )#define min( x, y )  ( ((x)   

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