uva 1385 - Billing Tables(字典树)

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题目链接：uva 1385 - Billing Tables

题目大意：给定n个电话前缀，每个前缀是一个区域的前缀，现在要生成一个新的电话单，即对于每个电话号码，从旧的电话单上从前向后遍历，如果出现前缀匹配，则该电话号码对应的即为当前的区号，要求生成的新电话单尽量小。

解题思路：用dfs建立字典树，在区间范围内的点对应均为对应的区号，注意如果70、71、72、...79都为SB的话，那么可以合并成7，并且对应区号为SB。
注意合并的条件为区号相同即可，并不是说对应旧电话单匹配位置相同。
注意这组数据：0 - 9 all

#include <cstdio>#include <cstring>#include <vector>#include <string>#include <map>#include <iostream>#include <algorithm>using namespace std;const int sigma_size = 10;typedef pair<string, string> pii;struct Node {    int val;    Node* next[sigma_size];    Node() {         val = 0;        memset(next, 0, sizeof(next));    }};void clear(Node* &p);void insert (Node* &p, int d, int L, int R, int tig);void dfs(Node* p, string str);int pushup(Node* &p, int tig);map<string, int> g;int n, m;string l, r, name[105];Node* root = NULL;vector<pii> vec;void init () {    vec.clear();    g.clear();    string tmp;    for (int i = 1; i <= m; i++) {        cin >> l >> tmp >> r >> name[i];        tmp = "";        int d = l.length() - r.length();        for (int i = 0; i < d; i++)            tmp += l[i];        tmp += r;        r = tmp;        if (l > r)            continue;        int k = i;        if (g.count(name[i]))            k = g[name[i]];        else            g[name[i]] = i;        insert(root, 0, 1, 1, k);    }}int main () {    int cas = 0;    while (cin >> m) {        if (cas++)            cout << endl;        init();        pushup(root, m+1);        dfs(root, "");        clear(root);        printf("%lu\n", vec.size());        for (int i = 0; i < vec.size(); i++)            cout << vec[i].first << " " << vec[i].second << endl;    }    return 0;}int pushup (Node* &p, int tig) {    if (p == NULL) {        p = new Node;        return p->val = tig;    }    if (p->val)        tig = p->val;    int k = pushup(p->next[0], tig);    for (int i = 1; i < sigma_size; i++) {        if (k != pushup(p->next[i], tig))            k = 0;    }    return p->val = k;}void dfs (Node* p, string str) {    if (p != root && p->val) {        if (p->val <= m && name[p->val] !=  "invalid")            vec.push_back(make_pair(str, name[p->val]));        return ;    }    for (int i = 0; i < sigma_size; i++) {        if (p->next[i] != NULL) {            char ch = '0' + i;            dfs(p->next[i], str + ch);        }    }}void insert (Node* &p, int d, int L, int R, int tig) {    if (p == NULL)        p = new Node;    if (p->val)        tig = p->val;    if (d >= l.length()) {        p->val = tig;        return;    }    if (L == 0 && R == 0) {        p->val = tig;        return;    } else if (L == 0) {        insert(p->next[r[d]-'0'], d + 1, 0, 1, tig);        for (int i = 0; '0' + i < r[d]; i++)            insert(p->next[i], d + 1, 0, 0, tig);    } else if (R == 0) {        insert(p->next[l[d]-'0'], d + 1, 1, 0, tig);        for (int i = l[d] - '0' + 1; i < sigma_size; i++)            insert(p->next[i], d + 1, 0, 0, tig);    } else if (r[d] == l[d]) {        insert(p->next[l[d]-'0'], d + 1, 1, 1, tig);    } else {        insert(p->next[l[d]-'0'], d + 1, 1, 0, tig);        insert(p->next[r[d]-'0'], d + 1, 0, 1, tig);        for (int i = l[d] + 1; i < r[d]; i++)            insert(p->next[i-'0'], d + 1, 0, 0, tig);    }}void clear(Node* &p) {    for (int i = 0; i < sigma_size; i++) {        if (p->next[i] != NULL)            clear(p->next[i]);    }    delete p;    p = NULL;}

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