IMMEDIATE DECODABILITY
Time Limit:1000MSMemory Limit:10000K
Total Submissions:11585Accepted:5517
Description
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:0000but 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 standard 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
0110001000009011001000009
Sample Output
Set 1 is immediately decodableSet 2 is not immediately decodable
题意：
判断二进制码是否能唯一识别，也就是一个二进制码不是另一个二进制码的前缀
题解：
最近在学字典树，搜的字典树的简单题做的，其实还可以排序做，先说字典树吧，在树的节点上做个标记，标记作用为是否是为一个二进制码的结尾，于是就有两种情况，，一种为在此码是前面出现的二进制码的前缀，这时只需要判断此码的结尾字符在树上存在，还有就是前面出现的二进制码是此码的前缀，这时只要判断标记就行了，注意初始化。
排序的方法，对二进制码按字典序排序，一个二进制码的前缀肯定在其前一个位置出现，具体原因，自己想下就很容易理解了O(∩_∩)O~~
字典树代码：
#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn=110000;struct node{int sign;int next[2];} a[maxn];int n=0;bool insert(char *s){int len,temp;len=strlen(s);int p=0;for(int i=0; i<len; i++){temp=s[i]-'0';if(a[p].next[temp]!=-1){p=a[p].next[temp];if(i==len-1)return 1;if(a[p].sign==1)return 1;}else{a[p].next[temp]=++n;p=n;}if(i==len-1)a[p].sign=1;}return 0;}int main(){char s[15];int count=1;while(~scanf("%s",s)){n=0;if(s[0]=='9')break;memset(a,-1,sizeof(a));int ans=0;if(insert(s)){ans=1;}while(scanf("%s",s)&&s[0]!='9'){if(insert(s)){ans=1;}}if(ans)printf("Set %d is not immediately decodable\n",count++);elseprintf("Set %d is immediately decodable\n",count++);}}
排序代码：
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;struct node{char s[15];} a[11000];bool operator < (node const& x,node const& y){return strcmp(x.s,y.s)<0;}int main(){int n=0;int count=1;while(~scanf("%s",a[n].s)&&a[n].s[0]!='9'){n++;while(scanf("%s",a[n].s)&&a[n].s[0]!='9'){n++;}sort(a,a+n);int ans=0;for(int i=1; i<n; i++){int len1=strlen(a[i-1].s);int len2=strlen(a[i].s);if(len1<=len2){int j;for(j=0; j<len1; j++)if(a[i-1].s[j]!=a[i].s[j])break;if(j==len1)ans=1;}}if(ans)printf("Set %d is not immediately decodable\n",count++);elseprintf("Set %d is immediately decodable\n",count++);n=0;}return 0;}
我知道我不是一个很好的记录者，但我比任何人都喜欢回首自己来时的路，
