字典树的数组实现 理解 + 类模板

模板

class Trie { 
public:
    static const int MAXN = 2e6+10;
    int tr[MAXN][30], sum[MAXN], tot = 0; 
    void insert ( char *s ) {
        register int len = strlen( s ), rt = 0; 
        for ( register int i = 0;i < len; ++i ) {
            register int id = s[i]-'a';
            if ( !tr[rt][id] ) tr[rt][id] = ++tot;
            sum[tr[rt][id]]++;
            rt = tr[rt][id];
        }
    }
    long long search ( char *s ) {
        long long res = 0; 
        register int len = strlen( s ), rt = 0; 
        for ( register int i = 0; i < len; ++i ) {
            register int id  = s[i]-'a';
            if ( !tr[rt][id] ) return 0;
            rt = tr[rt][id];
        }
        return sum[rt];
    }
}tr;

例题

HDU_1251 统计难题 ( map解法 和 字典树 )

理解

输出每次插入一个字符串的二维数组, 然后跟着模拟一遍, 这个二维数组存的是结点数, ( 字典树的节点是按插入顺序来增加 ), 和手绘的树其实一样. 注意 tot 其实就是结点数, 每次存入二维数组里, rt 是用来换行的也就是树的高度. 特别注意每次插入一个新的字符串的时候, rt 是从零开始. 和手绘的字典树图是一样的, 后插入的一个字符串如果在字典树里有相同的前缀, 会沿着这一条分支, 一直到不同的那个字符的节点位置在分叉.
这个二维数组里存的就是节点编号 !
根据不同的题意, 通过结点编号做索引来使用字典树的性质
字典树的数组实现 理解 + 类模板_第1张图片

banana
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
band
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0
4 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
bee
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0
4 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
absolute
10 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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0 0
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0
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0 0
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0 0 0 0 17 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0
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acm
10 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0
2 0 0 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0
0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 00
4 0 0 7 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 06 0
0 0 0 0 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0
6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 11 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12 00 0 0 0 0 0 0 0 0 0 0 0 0 0 13 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 14 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 16
0 0 0 0 17 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 19 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

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