hdu1305——Immediate Decodability

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.
Examples: Assume an alphabet that has symbols {A, B, C, D}
The following code is immediately decodable:
A:01 B:10 C:0010 D:0000
but this one is not:
A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)
Input
Write a program that accepts as input a series of groups of records from input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).
Output
For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.
Sample Input
01
10
0010
0000
9
01
10
010
0000
9
Sample Output
Set 1 is immediately decodable
Set 2 is not immediately decodable

也是找出字符串中是否有一个字符串是另一个字符串的前缀

代码:

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
bool flag;
char ph[20];
struct Node{
    //以该点结尾的单词数
    int c;
    int f;
    Node* Next[2];
    Node(){
        c=0;
        f=0;
        for(int i=0;i<2;i++){
            Next[i]=NULL;
        }
    }
};
void insert(Node* &rt,char* s){
    int l=strlen(s);
    int val;
    Node* p=rt;
    for(int i=0;i<l;i++){
        val=s[i]-'0';
        if(p->Next[val]==NULL){
            p->Next[val]=new Node();
        }
        p=p->Next[val];
        if(p->f){
            flag=true;
        }
    }
    p->f=1;
}
void clear(Node* root){
    if(root==NULL){
        return;
    }
    for(int i=0;i<2;i++){
        clear(root->Next[i]);
    }
    delete(root);
}
int main(void){
    int c=1;
    Node* root=new Node();
    while(~scanf("%s",ph)){
        if(ph[0]!='9'){
            insert(root,ph);
        }
        else{
            if(flag){
                printf("Set %d is not immediately decodable\n",c++);
            }
            else{
                printf("Set %d is immediately decodable\n",c++);
            }
            flag=false;
            clear(root);
            root=new Node();
        }
    }
    return 0;
}
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程序员牛肉:主要是因为小厂的资金本来就很吃紧,所以更喜欢有实习经历的同学。来了就能上手。 而大厂因为钱多,实习生一天三四百的就不算事。所以愿意培养你,在面试的时候也就不在乎你有没有实习(除非是同级别大厂的实习。) 按照你的简历来看,同质化太严重了。项目也很烂大街。 要么换项目,要么考研。 你现在选择工作的话,前景不是很好了。
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07-02 13:50
闽江学院 Java
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