题目链接：hdu 4760 Good Firewall
题目大意：有1个防火墙，具有添加1个子网络，删除1个子网络，和转发包的操作。
解题思路：对子网掩码前缀建立字典树，每一个前缀终止节点用1个set记录属于哪些子网络，ip下限。那末增加和删除操
作既可以解决了。对查询操作，分别查询两个ip，处理除两个ip可能属于的网络，判断有没有共同便可。
#include <cstdio>
#include <cstring>
#include <set>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<int, ll> pii;
typedef set<pii>::iterator iter;
const int maxn = 1024 * 15 + 10;
const int maxm = 105;
const int sigma_size = 2;
struct Tire {
int sz, sv;
int g[maxn * maxm][sigma_size];
int idx[maxn * maxm], cnt[1030];
set<pii> ans, vis[maxn];
void init();
int newSet();
void addSet(int id, ll limt);
void insert(char* str, pii x);
void remove(char* str, pii x);
void find(char* str);
bool judge(char* a, char* b);
}T;
struct Network {
int n;
ll suf[20];
char ip[20][maxm];
Network() {
n = 0;
memset(suf, 0, sizeof(suf));
}
}net[1030];
ll change(char* ans, bool flag) {
char str[105];
int n = strlen(str), a[4], b;
if (flag)
scanf("%d.%d.%d.%d/%d", &a[0], &a[1], &a[2], &a[3], &b);
else {
scanf("%d.%d.%d.%d", &a[0], &a[1], &a[2], &a[3]);
b = 32;
}
int t = 0;
ll ret = 0;
for (int i = 0; i < 4; i++) {
for (int j = 7; j >= 0; j--) {
if (t < b)
ans[t] = ((a[i]>>j)&1) + '0';
else if (((a[i]>>j)&1))
ret |= (1LL<<(31-t));
t++;
}
}
ans[t] = ' ';
return ret;
}
int main () {
int id;
char op[5], a[maxm], b[maxm], c[maxm];
T.init();
while (scanf("%s", op) == 1) {
if (op[0] == 'E') {
scanf("%d", &id);
scanf("%d", &net[id].n);
for (int i = 0; i < net[id].n; i++) {
net[id].suf[i] = change(net[id].ip[i], a);
T.insert(net[id].ip[i], make_pair(id, net[id].suf[i]));
}
} else if (op[0] == 'D') {
scanf("%d", &id);
for (int i = 0; i < net[id].n; i++)
T.remove(net[id].ip[i], make_pair(id, net[id].suf[i]));
net[id].n = 0;
} else {
change(a, 0);
change(b, 0);
printf("%c
", T.judge(a, b) ? 'F' : 'D');
}
}
return 0;
}
bool Tire::judge(char* a, char* b) {
memset(cnt, 0, sizeof(cnt));
T.find(a);
for (iter i = ans.begin(); i != ans.end(); i++)
cnt[i->first] = 1;
T.find(b);
for (iter i = ans.begin(); i != ans.end(); i++)
if (cnt[i->first])
return true;
return false;
}
void Tire::init() {
sz = sv = 1;
idx[0] = 0;
vis[0].clear();
memset(g[0], 0, sizeof(g[0]));
}
int Tire::newSet() {
vis[sv].clear();
return sv++;
}
void Tire::addSet(int id, ll limt) {
for (iter i = vis[id].begin(); i != vis[id].end(); i++)
if (i->second <= limt)
ans.insert(*i);
}
void Tire::insert(char* str, pii x) {
int u = 0, n = strlen(str);
for (int i = 0; i < n; i++) {
int v = str[i] - '0';
if (g[u][v] == 0) {
idx[sz] = 0;
memset(g[sz], 0, sizeof(g[sz]));
g[u][v] = sz++;
}
u = g[u][v];
}
if (idx[u] == 0)
idx[u] = newSet();
vis[idx[u]].insert(x);
}
void Tire::remove(char* str, pii x) {
int u = 0, n = strlen(str);
for (int i = 0; i < n; i++) {
int v = str[i] - '0';
if (g[u][v] == 0) {
idx[sz] = 0;
memset(g[sz], 0, sizeof(g[sz]));
g[u][v] = sz++;
}
u = g[u][v];
}
vis[idx[u]].erase(x);
}
void Tire::find(char* str) {
ans.clear();
ll ret = 0;
int u = 0, n = strlen(str);
for (int i = 0; i < n; i++) {
int v = str[i] - '0';
if (g[u][v] == 0)
return;
u = g[u][v];
if (idx[u]) {
ll ret = 0;
for (int j = i+1; j < n; j++)
if (str[j] == '1')
ret |= (1LL<<(31-j));
addSet(idx[u], ret);
}
}
}