Problem I. International Collegiate Routing ContestGym

题目意思是，说给予n个32位IP地址，每个地址的后面有一个K，每个K代表与当前所给的IP地址的前K位相同的话，就为同一IP地址，然后问你，最少再补充多少的IP地址使得能构成所有的IP地址。

 思路：字典树的题目，对于每个32位的串，输入字典树中，如果当前K小于32那么就只导入到第K位，然后在当前节点做标记，建完树后，跑一遍DFS就可以了。

#include <iostream>#include <stdio.h>#include <algorithm>#include <cstring>#include <queue>using namespace std;#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct node{    int next[2];    int v;    void init(){        v=0;        memset(next,-1,sizeof(next));    }};struct node L[4000000];int tot=0;void add(char a[],int len, int x){    int now=0;    for(int i=0;i<len;i++){        int tmp=a[i]-'0';        int next=L[now].next[tmp];        if (i + 1 == x) {        L[now].next[tmp] = -2;        return;        }        if (next == -2) {        return;        }        if(next==-1){            next=++tot;            L[next].v=-1;            L[next].init();            L[now].next[tmp]=next;        }        now=next;        L[now].v++;    }}int T, n;int a,b,c,d,k;char s[50];int  becom_cha(int x, int len){int l = len;int r = len;for (int i = 0; i < 7; i++) {s[r++] = x%2 + '0';x = x/2;}s[r++] = x+'0';reverse(s+l, s+r);return r;}struct point{long long int x, y;point (long long int a = 0, long long int b = 0) {x = a;y = b;}};int a1[100];vector<point> q;void dfs(int now, long long int x, int len){if (len == 32) return;for (int i = 0; i < 2; i++) {if (L[now].next[i] == -2) continue;else if (L[now].next[i] == -1) {long long int ans = 2*x+i;for (int j = len+1; j < 32; j++) {ans = 2 * ans;}q.push_back(point(ans, len+1));}else dfs(L[now].next[i], x*2+i, len+1);}return;}int main(){scanf("%d", &T);int cas = 0;while (T--) {cas++;scanf("%d", &n);tot = 0;L[tot].init();if (n == 0) {printf("Case #%d:\n1\n0.0.0.0/0\n", cas);continue;}for (int i = 1; i <= n; i++) {int len = 0;scanf("%d.%d.%d.%d", &a, &b, &c, &d);char ch=getchar();if(ch=='/')scanf("%d",&k);else k=0;len = becom_cha(a, len);len = becom_cha(b, len);len = becom_cha(c, len);len = becom_cha(d, len);add(s, len, k);}q.clear();dfs(0, 0, 0);printf("Case #%d:\n", cas);if(k==0){printf("0\n");continue;}printf("%d\n",q.size());for (int i = 0; i < q.size(); i++) {long long int ans = q[i].x;for (int j = 0; j < 3; j++) {a1[j] = ans%256;ans /= 256;}a1[3] = ans;for (int j = 3; j >= 0; j--) {printf("%d%c", a1[j],j==0?'/':'.');}printf("%lld\n", q[i].y);}}}