HDOJ 1298 T9（trie树简单应用）

题意：

一串数字，给出出现频率最高的那个单词。

思路：

字典树：首先是建树，其实最麻烦的是概率处理部分，因为题目给定的单词都是按照字典有序排列的，所以可以对于概率进行一遍预处理，然后建树的时候就可以直接更新判断了。

题目同 POJ 1451 http://poj.org/problem?id=1451

#include 
#include 
using namespace std;struct node {bool isword;int prob, index;int child[10];node(bool _isword, int _prob, int _index) :   isword(_isword), prob(_prob), index(_index){ memset(child, 0, sizeof(child)); }
};vector<node> trie;int tab[256];
char dict[1010][102];
int prob[1010][102];void Insert(const char* word, int i, int index, int rt)
{if (word[i] == '\0'){trie[rt].isword = true;return ;}int m = tab[word[i]];if (trie[rt].child[m]){if (trie[trie[rt].child[m]].prob < prob[index][i]){trie[trie[rt].child[m]].prob = prob[index][i];trie[trie[rt].child[m]].index = index;}else if (trie[trie[rt].child[m]].prob == prob[index][i]){if (strncmp(dict[index], dict[trie[trie[rt].child[m]].index], i + 1) < 0)trie[trie[rt].child[m]].index = index;}Insert(word, i + 1, index, trie[rt].child[m]);}else{trie.push_back(node(false, 0, 0));trie[rt].child[m] = trie.size() - 1;trie[trie[rt].child[m]].index = index;trie[trie[rt].child[m]].prob = prob[index][i];Insert(word, i + 1, index, trie[rt].child[m]);}
}int Query(const char* word, int rt)
{if (*word == '\0')return trie[rt].index;int m = *word - '0';if (!trie[rt].child[m])return -1;elsereturn Query(word + 1, trie[rt].child[m]);
}void Init()
{tab['a'] = tab['b'] = tab['c'] = 2;tab['d'] = tab['e'] = tab['f'] = 3;tab['g'] = tab['h'] = tab['i'] = 4;tab['j'] = tab['k'] = tab['l'] = 5;tab['m'] = tab['n'] = tab['o'] = 6;tab['p'] = tab['q'] = tab['r'] = tab['s'] = 7;tab['t'] = tab['u'] = tab['v'] = 8;tab['w'] = tab['x'] = tab['y'] = tab['z'] = 9;
}int main()
{int cases;Init();scanf("%d", &cases);for (int c = 1; c <= cases; ++c){int n, p;scanf("%d", &n);printf("Scenario #%d:\n", c);trie.clear();trie.push_back(node(false, 0, 0));memset(prob, 0, sizeof(prob));for (int i = 1; i <= n; ++i){scanf("%s %d", dict[i], &p);int len = strlen(dict[i-1]);for (int j = 0; dict[i][j] != '\0'; ++j){prob[i][j] = p;if (j < len && !strncmp(dict[i], dict[i-1], j + 1))prob[i][j] += prob[i-1][j];}Insert(dict[i], 0, i, 0);}char word[110], is[110], os[110];scanf("%d", &n);while (n--){scanf("%s", is);for (int i = 0; is[i] != '1'; ++i){word[i] = is[i];word[i+1] = '\0';int id = Query(word, 0);if (id != -1){int j = 0;for (j = 0; j <= i; ++j)os[j] = dict[id][j];os[j] = '\0';printf("%s\n", os);}else printf("MANUALLY\n");}printf("\n");}printf("\n");}return 0;
}

转载于:https://www.cnblogs.com/kedebug/archive/2013/01/24/2875545.html

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