IMMEDIATE DECODABILITY

題目：IMMEDIATE DECODABILITY

題解：先按照每個字符串的長短從小到大排序，然後依次插入每一個字符串，並且每一個插完之後，做一下標記，如果下一個字符串訪問到這個標記，則說明Not

#include <string>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int N = 1e5+10;
const int inf = 0x7FFFFFFF;

int tree[N][2];
int cnt = 1;
string str[N];

bool insert(string s){
    int n = s.size(),p = 1;
    for(int i = 0;i < n;i++){
        int x = s[i]-48;
        if(!tree[p][x]) tree[p][x] = ++cnt;
        p = tree[p][x];
        if(tree[p][0]==-1&&tree[p][1]==-1) return true;
    }
    tree[p][0] = tree[p][1] = -1;
    return false;
}

void traver(){
    for(int i = 1;i <= 15;i++){
        printf("%d:%d %d\n",i,tree[i][0],tree[i][1]);
    }
    cout<<endl;
}
bool cmp(string a,string b){
    return a.size() < b.size();
}
void solve(int n){
    sort(str,str+n,cmp);
    
    for(int i = 0;i < n;i++){
        if(insert(str[i])){
            printf("is not immediately decodable\n");
            return;
        }
        /*
        cout<<str[i]<<endl;
        traver();
        */
    }
    printf("is immediately decodable\n");
}
int main(){
    string s;
    int len = 1;
    int n = 0;
    while(cin >> s){
        if(s[0] == '9'){
            printf("Set %d ",len++);
            solve(n);
            memset(tree,0,sizeof tree);
            n = 0;
            cnt = 1;
        }else str[n++] = s;
    }
    return 0;
}