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字典树 Trie （HDU 1671）

2024-07-07 10:28:21   231人阅读

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
2
3
911
97625999
91125426
5
113
12340
123440
12345
98346
 

Sample Output
NO
YES

题意 ： 给出n个字符串 ， 且任意两个字符串不存在前缀关系，那么输出yes， 否则输出no 。

思路 ： 建立Trie ，val[]的值代表字符串是否为结束点 ，非0 代表是，0代表不是 。 然后再遍历n个字符串 ， 计算每一个字符串的val[]，统计出val[]的个数，假如val[]非0的个数大于1，说明这个串存在前缀，返回不满足。


#include <string.h>
#include <stdio.h>
int ch[100050][11];
int val[100050];
int sz;
int idx(char c) { return c-'0'; }

void inser(char *s , int v) {
    int u=0 , n=strlen(s);
    for(int i=0;i<n;i++) {
        int c=idx(s[i]);
        if( !ch[u][c] ) {
        //    memset(ch[sz] ,0 ,sizeof(sz));
            val[sz]=0;
            ch[u][c]=sz++;
        }
        u=ch[u][c];
    }
    val[u]=v;
}

int Find(char *s ) {
    int i, n=strlen(s) , u=0 ,cnt=0;
    for(i=0;i<n;i++) {
        int c=idx(s[i]);
        u=ch[u][c];
        if(val[u] != 0) cnt++;
    }
    if(cnt>1) return 0;
    else return 1;
}

char a[10005][11];

int main()
{
    int T,n,i;
    scanf("%d%*c",&T);
    while(T--) {
        sz=1;
        memset(val,0,sizeof(val));
        memset(ch,0,sizeof(ch));
        scanf("%d%*c",&n);
        for(i=0; i<n;i++) {
            scanf("%s",a[i]);
            inser( a[i], 1 );
        }
        for(i=0;i<n;i++) {
            if(!Find(a[i])) {
                printf("NO\n");
                break;
            }
        }
        if(i == n) printf("YES\n");
    }
    return 0;
}


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