題目連結：這裡

題意：輸入IPv4地址空間中的一些子網號，構成一個網路集合。
輸出最小的一個網路集合，要求其與輸入集合沒有交集，且相對IPv4地址空間全集，是輸入集合的補集。輸出集合包含的子網號，格式遵循網路規範
做法：tire樹，把所有輸入的子網插入trie裡面，然後遍歷一下，如果一個結點的左兒子和右兒子有一個是標記過的，就輸出沒有標記的那個兒子的子網。

#include <bits/stdc++.h>

using namespace std;

#define pii pair<int, int>
#define MP make_pair
#define LL long long
 

#define ls (i << 1)
#define rs (ls | 1)
#define md (ll + rr >> 1)
#define lson ll, md, ls
#define rson md + 1, rr, rs
#define eps 1e-8
#define inf 0x3f3f3f3f
#define mod 1000000007
#define N 100010
#define M 200020
#define fi first
#define se second

char buf[32000000],*pt = buf,*o = buf;
int getint(){
    int
 
 f = 1,x = 0;
    while((*pt != '-') && (*pt < '0' || *pt > '9'))    pt ++;
    if(*pt == '-')    f = -1,pt ++;    else    x = *pt++ - 48;
    while(*pt >= '0' && *pt <= '9')    x = x * 10 + *pt ++ - 48;
    return x * f;
}
char getch(){
    char ch;
    while(*pt < 'A' || *pt
 
 > 'Z')    pt ++;
    ch=*pt;pt++;
    return ch;
}

int ch[N*30][2], cnt, tot, n;
bool vis[N*30];
pair<LL, int> p[N];
int insert(LL v, int k){
    int u = 1;
    for(int i = 31; i >= 0 && k; --i, --k){
        int c = v >> i & 1;
        if(!ch[u][c]){
            ch[u][c] = ++tot;
            memset(ch[tot], 0, sizeof ch[tot]);
            vis[tot] = 0;
        }
        u = ch[u][c];
    }
    vis[u] = 1;
}
void print(LL v, int dep){
    v <<= dep;
    int a = v >> 24;
    v ^= (LL)a << 24;
    int b = v >> 16;
    v ^= (LL)b << 16;
    int c = v >> 8;
    v ^= (LL)c << 8;
    int d = v;
    printf("%d.%d.%d.%d/%d\n", a, b, c, d, 32 - dep);
}
int query(int u, LL x, int dep){
    if(u == 0 || vis[u]) return vis[u];
    if(dep == -1)
        return vis[u];
    int p1 = query(ch[u][0], x << 1, dep - 1);
    int p2 = query(ch[u][1], x << 1 | 1, dep - 1);
    if(p1 && !p2){
        p[cnt++] = MP(x << 1 | 1, dep);
    }
    else if(!p1 && p2){
        p[cnt++] = MP(x << 1, dep);
    }
    return p1 | p2;
}

int main(){
    int cas, kk = 0;
    scanf("%d", &cas);
    while(cas--){
        scanf("%d", &n);
        tot = 1;
        memset(ch[tot], 0, sizeof ch[tot]);
        memset(ch[0], 0, sizeof ch[0]);
        vis[0] = vis[1] = 0;
        for(int i = 0; i < n; ++i){
            int a, b, c, d, k;
            scanf("%d.%d.%d.%d/%d", &a, &b, &c, &d, &k);
            LL v = 1LL * a * (1 << 24) + b * (1 << 16) + c * (1 << 8) + d;
            insert(v, k);
        }
        printf("Case #%d:\n", ++kk);
        if(n == 0){
            printf("1\n0.0.0.0/0\n"); continue;
        }
        cnt = 0;
        query(1, 0LL, 31);
        printf("%d\n", cnt);
        for(int i = 0; i < cnt; ++i)
            print(p[i].fi, p[i].se);

    }
    return 0;
}