hihoCoder 403 Forbidden 字典树

题意：给定n个规则，m个ip，问这些ip是否能和某个规则匹配，如果有多个规则，则匹配第一个。如果没能匹配成功，则认为是”allow”，否则根据规则决定是”allow”或者”deny”.

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思路：字典树，将所有ip全部转换为01串，在字典树上面插入查找。
在求二进制时，注意补前缀0。
坑点：
1.字典树上每一个规则都应该带一个序号，表示这是第几个规则，因为题目要求第一个匹配，那么当你利用ip去寻找匹配的规则时，应该是找到序号最小的。
2.出现deny 0.0.0.0/0，说明所有无法匹配ip的都是deny，因为这些ip和这条规则匹配。
给大家贴几个测试数据:
5 2
deny 0.0.0.0/0
allow 0.0.0.0/0
deny 0.0.0.0/1
deny 0.0.0.0/2
allow 123.234.12.23/3
123.234.12.23
0.234.12.23
答案: NO NO
2 1
deny 1.1.1.1
allow 127.0.0.1
1.1.1.222
答案：YES

AC代码

#include <cstdio>#include <cmath>#include <cctype>#include <bitset>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>using namespace std;#pragma comment(linker, "/STACK:1024000000,1024000000") #define eps 1e-10#define inf 0x3f3f3f3f#define PI pair<int, int> typedef long long LL;const int maxn = 1e5 + 5;bool is_zero, tag;struct node{    int ok, order;    node *nex[2];    node() {        ok = -1; //从未访问过         nex[0] = nex[1] = NULL;    }}*root;void init() {    root = new node();}void insert(string &s, int n, int ok, int order) {    node *p = root, *q;    for(int i = 0; i < n; ++i) {        int u = s[i] - '0';        if(p->nex[u] == NULL) {            q = new node();            p->nex[u] = q;        }        p = p->nex[u];        if(i == n-1 && p->ok == -1) {            p->ok = ok;            p->order = order;        }     }}bool search(string &s) {    node *p = root;    bool ok = true;    int order = inf;    for(int i = 0; i < s.size(); ++i) {        int u = s[i]-'0';        if(p->nex[u] == NULL) break;        else {            if(p->nex[u]->ok != -1 && p->nex[u]->order < order) {                order = p->nex[u]->order;                ok = p->nex[u]->ok;            }        }        p = p->nex[u];    }    //printf("%d\n", order);    if(is_zero && order == inf) return tag;    return ok;}string get_binary(int x) {    stack<int>sta;    do{        sta.push(x%2);        x /= 2;    }while(x);    string ans = "";    //补前缀0     for(int i = 0; i < 8 - sta.size(); ++i) {        ans += '0';    }    //得到二进制    while(!sta.empty()) {        ans += sta.top() + '0';        sta.pop();    }     return ans;}void deal(char *s, int ok, int order) {    int len = strlen(s);    int ind = -1;    for(int i = 0; i < len; ++i) {        if(s[i] == '/') {            ind = i;            break;        }    }    string ip = "";    if(ind == -1) ind = len;    for(int i = 0; i < ind;) {        if(s[i] >= '0' && s[i] <= '9') {            int num = 0;            while(i < ind && s[i] >= '0' && s[i] <= '9'){                num = num * 10 + (s[i] - '0');                ++i;            }            ip += get_binary(num);        }        else ++i;    }    int n = 32;    if(ind != -1 && ind != len) {        int num = 0;        for(int i = ind+1; i < len; ++i) {            num = num * 10 + s[i] -'0';        }        if(num == 0 && !is_zero) {            is_zero = 1;            tag = ok;        }         n = num;    }    insert(ip, n, ok, order);}int main() {    init();    char kind[10], ip[40];    int n, m;    while(scanf("%d%d", &n, &m) == 2) {        is_zero = 0;        for(int i = 0; i < n; ++i) {            scanf("%s %s", kind, ip);            int ok = 0;            if(kind[0] == 'a') ok = 1;            deal(ip, ok, i);        }        for(int i = 0; i < m; ++i) {            scanf("%s", ip);            int x1, x2, x3, x4;            sscanf(ip, "%d.%d.%d.%d", &x1, &x2, &x3, &x4);            string res = get_binary(x1) + get_binary(x2) + get_binary(x3) + get_binary(x4);            if(search(res)) printf("YES\n");            else printf("NO\n");        }    }    return 0;} 

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