UVa OJ
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)
The Sample Input describes the examples above.
01 10 0010 0000 9 01 10 010 0000 9
Set 1 is immediately decodable Set 2 is not immediately decodable题目大意就是每一个code都不能是其他code的前缀。
可以先通过排序,按照code的位数从小到大排序(这里有点小技巧,因为二维数组是不能直接排序的,可以通过改为结构体,或是用指针排序)
sort对二维字符数组排序
http://blog.csdn.net/unimen/article/details/6843977
#include<iostream> #include<cstdio> #include<cstring> #include<ctype.h> #include<algorithm> using namespace std; struct node { char str[10]; int x; }; bool cmp(node s, node v) { return s.x<v.x; } node code[10000]; int main () { int t=0,i,j,p=1; while(gets(code[t].str)) { bool isID=true; if (code[t].str[0]=='9') { sort(code,code+t,cmp); for (i=0; i<t; i++) { for (j=i+1; j<=t; j++) { int len=strlen(code[i].str),k; for (k=0; k<len; k++) if (code[i].str[k]!=code[j].str[k]) break; if (k==len) { isID=false; break; } } if (!isID) break; } if (isID) printf("Set %d is immediately decodable\n",p); else printf("Set %d is not immediately decodable\n",p); t=0; memset(code,0,sizeof(code)); p++; } else { code[t].x=strlen(code[t].str); t++; } } return 0; }