Uva OJ 644 - Immediate Decodability

1Y的一道题

#include<string.h>#include<stdio.h>#include<stdlib.h>int cmp(const void *a,const void *b){char *p=(char *)a;char *q=(char *)b;return strcmp(p,q);}int f(char *a,char *b){if(strlen(b)<strlen(a))return 0;int i,len=strlen(a);for(i=0;i<len;i++){if(a[i]!=b[i])return 0;}return 1;}int main(){char map[20][20];char temp[20];int i;int pm=0;int time=0;while(scanf("%s",temp)!=EOF){int flag=0;if(strlen(temp)==1&&temp[0]=='9')flag=1;if(!flag){for(i=0;i<=strlen(temp);i++)map[pm][i]=temp[i];pm++;}else {flag=0;qsort(map,pm,sizeof(map[0]),cmp);int orz=0;for(i=1;i<pm;i++){if(f(map[i-1],map[i]))orz=1;if(orz)break; }if(orz)printf("Set %d is not immediately decodable\n",++time);else printf("Set %d is immediately decodable\n",++time);pm=0;}}return 0;}

数据规模很小,大点的话估计要用树了==

排序,判断相邻是否为前缀关系