Find the Clones(字典树之哈希功能)

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萌萌哒的链接:http://poj.org/problem?id=2945

题目的意思就是找每一个字符串出现的次数,输出出现1-n次的字符串的个数.

字典树的哈希应用.index记录相同字符串出现的次数,最后dfs查找. 因为题目中只有4个字母,所以dfs总的时间复杂度

为4*字典树的总结点个数.

字典树用链表写挺方便的,就是内存花销太大,比如这道题目如果把内存开成26,就MLE了.

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <set>#include <map>#include <queue>#include <vector>#include <cstdlib>#include <algorithm>#define ls u << 1#define rs u << 1 | 1#define lson l, mid, u << 1#define rson mid + 1, r, u << 1 | 1#define INF 0x3f3f3f3f#define MAX 4using namespace std;typedef long long ll;const int M = 2e4 + 100;const int mod = 2147483647;char s[25];int num[M],d[] = {'A'-'A','C'-'A','T'-'A','G'-'A'};struct Trie {    int index;    Trie *next[MAX];    Trie() {        index = 0;        memset(next,0,sizeof(next));    }};void Trie_insert(Trie *tr,int len) {    if(s[len]) {        int u = s[len] - 'A';        for(int i = 0; i < 4; i++){            if(u == d[i]){                u = i;                break;            }        }        if(s[len + 1] == '\0' && tr->next[u]) {            tr->next[u]->index++;            return ;        }        if(tr->next[u] == 0) tr->next[u] = new Trie;        Trie_insert(tr->next[u],len + 1);    } else {        tr->index = 1;    }}void dfs(Trie *tr) {    if(tr->index) num[tr->index]++;    for(int u = 0; u < 4; u++) {        if(tr->next[u]) dfs(tr->next[u]);    }}void deal_Trie(Trie *tr){    if(tr == NULL) return ;    for(int u = 0; u < 4; u++){        if(tr->next[u]) dfs(tr->next[u]);    }    free(tr);}int main() {    int n,m;    while(~scanf("%d %d",&n,&m) && (n | m)) {        Trie *root = new Trie;        memset(num,0,sizeof(num));        for(int i = 0; i < n; i++) {            scanf("%s",s);            Trie_insert(root,0);        }        dfs(root);        for(int i = 1; i <= n; i++) {            printf("%d\n",num[i]);        }        //deal_Trie(root);    }    return 0;}

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