Phone List

题目链接

Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11757    Accepted Submission(s): 3995

Problem Description

Given a list of phone numbers, determine if it is consistent in the sense that no number is the prefix of another. Let’s say the phone catalogue listed these numbers:
1. Emergency 911
2. Alice 97 625 999
3. Bob 91 12 54 26
In this case, it’s not possible to call Bob, because the central would direct your call to the emergency line as soon as you had dialled the first three digits of Bob’s phone number. So this list would not be consistent.

 

Input

The first line of input gives a single integer, 1 <= t <= 40, the number of test cases. Each test case starts with n, the number of phone numbers, on a separate line, 1 <= n <= 10000. Then follows n lines with one unique phone number on each line. A phone number is a sequence of at most ten digits.

 

Output

For each test case, output “YES” if the list is consistent, or “NO” otherwise.

 

Sample Input

2391197625999911254265113123401234401234598346

 

Sample Output

NOYES

 数组代码：

/*Author: 2486Memory: 2764 KBTime: 157 MSLanguage: G++Result: Accepted*/#include <cstdio>#include <cstring>#include <vector>#include <cmath>#include <queue>#include <algorithm>using namespace std;const int MAXN = 10000000 + 5;struct node {    int v,next[10];    void init() {        v = -1;//表示不是字符串的结尾        memset(next, -1, sizeof(next));    }} L[MAXN];int tot, T, n;bool flag;void add(char a[], int len) {    int now = 0;    for(int i = 0; i < len; i ++) {        int tmp = a[i] - '0';        int next = L[now].next[tmp];        if(next == -1) {//增加尾部字符            next = ++ tot;            L[next].init();            L[now].next[tmp] = tot;        }        if(L[next].v == 0)flag = true;        now = next;    }    L[now].v = 0;//表示此时是字符串的结尾    for(int i = 0; i < 10; i ++) {        if(L[now].next[i] != -1)flag = true;//检查是否为其他串的子串    }}int main() {    char op[15];    scanf("%d", &T);    while(T --) {        scanf("%d", &n);        tot = 0;        flag = false;        L[0].init();        for(int i = 0; i < n; i ++) {            scanf("%s", op);            if(!flag)add(op,strlen(op));        }        printf("%s\n",flag ? "NO" : "YES");    }    return 0;}

指针代码：

#include <iostream>#include <string>#include <cstdlib>#include <cstring>#include <cstdio>char str[12];const int MAX=10;using namespace std;struct Tire{    Tire *next[MAX];    int v;    int vv;};Tire *root;void creatTire(char *str){int len=strlen(str);Tire *p=root,*q;for(int i=0;i<len;i++){int id=str[i]-'0';if(p->next[id]==NULL){    q=(Tire *)malloc(sizeof(Tire));    q->v=1;    for(int j=0;j<MAX;j++){        q->next[j]=NULL;    }    q->vv=0;    p->next[id]=q;    p=q;}else{    p->next[id]->v++;    p=p->next[id];}}p->vv=-1;}bool findTire(char *str){int len=strlen(str);Tire *q=root;for(int i=0;i<len;i++){    int id=str[i]-'0';    q=q->next[id];    if(q==NULL){        return false;    }    if(q->vv==-1){            if(q->v>1){                return true;            }else{            return false;            }    }}return true;}void delTire(Tire *to){int i;if(to==NULL){    return ;}for(i=0;i<MAX;i++){     delTire(to->next[i]);}free(to);return;}int main(){int t,n;bool success=false;//freopen("D://imput.txt","r",stdin);cin>>t;while(t--){    success=false;    root=(Tire *)malloc(sizeof(Tire));    for(int i=0;i<MAX;i++){        root->next[i]=NULL;        root->vv=0;        root->v=0;    }    cin>>n;    for(int i=0;i<n;i++){        cin>>str;        creatTire(str);    if(findTire(str)){        success=true;    }    }    if(!success){        cout<<"YES"<<endl;    }else{    cout<<"NO"<<endl;    }    delTire(root);}return 0;}