POJ 3630 - Phone List （Trie）

Phone List

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 21801 Accepted: 6779

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:

• Emergency 911
• Alice 97 625 999
• 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 withn, the number of phone numbers, on a separate line, 1 ≤ n ≤ 10000. Then followsn 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

Source

Nordic 2007

题意：

判断输入的电话号码中是否有电话号码是另外号码的前缀

思路：

Trie 但是不能直接插入判断　比如

2

1234

123

一种解决方法是按大小排序　然后依次插入判断

#include <stdio.h>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define S64I1(a) scanf(iform1, &(a))#define P64I1(a) printf(oform1, (a))#define FOR(i, s, t) for(int (i)=(s); (i)<(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxsize = 200000;struct Trie {    int size;    int val[maxsize];    int ch[maxsize][10];    Trie() {        clear();    }    void clear() {        size = 1;        memset(ch[0], 0, sizeof(ch[0]));    }    int idx(char c) {        return c - '0';    }    bool insert(const char * s, int v) {        int len = strlen(s);        int u = 0;        for(int i=0; i<len; i++) {            int c = idx(s[i]);            if(!ch[u][c]) {                memset(ch[size], 0, sizeof(ch[size]));                val[size] = 0;                ch[u][c] = size++;            }            if(val[u]) return false;            u = ch[u][c];        }        val[u] = v;        return true;    }};char s[20];Trie trie;string ss[10000+20];int main() {    int T;    scanf("%d", &T);    while(T--) {        int n;        trie.clear();        scanf("%d", &n);        for(int i=0; i<n; i++) {            cin>>ss[i];        }        sort(ss, ss+n);        int ans = 1;        for(int i=0; i<n; i++) {            if(ans && !trie.insert(ss[i].c_str(), 1)) {                ans = 0;            }        }        if(ans) puts("YES");        else puts("NO");    }    return 0;}

另外一种方法是给每个单词的路径标记一下　然后插入的时候判断一下就行了

#include <stdio.h>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define S64I1(a) scanf(iform1, &(a))#define P64I1(a) printf(oform1, (a))#define FOR(i, s, t) for(int (i)=(s); (i)<(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxsize = 200000;struct Trie {    int size;    int val[maxsize];    int ch[maxsize][10];    Trie() {        clear();    }    void clear() {        size = 1;        memset(ch[0], 0, sizeof(ch[0]));    }    int idx(char c) {        return c - '0';    }    bool insert(const char * s) {        int len = strlen(s);        int u = 0;        for(int i=0; i<len; i++) {            int c = idx(s[i]);            if(!ch[u][c]) {                memset(ch[size], 0, sizeof(ch[size]));                val[size] = 0;                ch[u][c] = size++;            }            if(val[u] == 2) return false;            val[u] = 1;            u = ch[u][c];        }        if(val[u]) return false;        val[u] = 2;        return true;    }};char s[100];Trie trie;int main() {    int T;    scanf("%d", &T);    while(T--) {        int n;        trie.clear();        scanf("%d", &n);        int ans = 1;        for(int i=0; i<n; i++) {            scanf("%s", s);            if(ans && !trie.insert(s)) {                ans = 0;            }        }        if(ans) puts("YES");        else puts("NO");    }    return 0;}