HDU 1671 字典树


Phone List

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22044    Accepted Submission(s): 7479


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


思路: 每次建边的时候进行查询. 不满足的情况有两种. 1.在还未建完边的时候遇到某个电话的末尾 2.在刚好建完边的时候发现自己这条边已经被建过了.
          做的时候少考虑一种情况 导致WA1.


源码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define rep(i, n) for(int i=0; i P;
const int INF = 0x7fffffff;
const int MAX_N = 4e4+5;
const int MAX_V = 0;
const int MAX_M = 0;
const int MAX_Q = 1e5+5;

void show(string a, int val){
	cout<nxt[id]!=NULL){
			p = p->nxt[id];
			if(i==str.size()-1)
				flag = false;
		}
		else{
			q = new Trie;
			p->nxt[id] = q;
			p = p->nxt[id];
		}
		if(p->is_end)
			flag = false;
	}
	p->is_end = true;
}

void clear(Trie *q){
	for(int i=0; i<10; i++)
		if(q->nxt[i]!=NULL)
			clear(q->nxt[i]);
	delete q;
}

int main(){
	ios::sync_with_stdio(0);
	int T; cin>>T;
	while(T--){
		for(int i=0; i<10; i++){
			if(root.nxt[i]!=NULL)
				clear(root.nxt[i]);
			root.nxt[i] = NULL;
		}
		string a; int n; cin>>n;
		flag = true;
		for(int i=0; i>a;
			if(flag) add_edge(a);
		}
		cout<<(flag?"YES\n":"NO\n");
	}
}



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