[UVA][字串理] 10438 - Meta Editor＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
49的鼓 6站台

2013-01-21 08:25:40| 人662| 回1 | 上一篇 | 下一篇

[UVA][字串理] 10438 - Meta Editor

推 0 收藏 0 0 站台

Brightness of Brain Contest
Problem J
Time limit: 1 second Memory: 16 MB

Meta Editor

The Problem

You got a job as a programmer in the ACM (A Cool Meta-editor) company, which has has been serving as a producer of one of the coolest text editor of the world. In the update process of the text editor program of that company you are assigned a job to make a program unit which will grammatically correct a line of text. Your job is very simple, take a line of text as input and produce a line of output without the words repeated.

The Input

Contains several lines of text. Each line contains several English words having no more then 50 characters each. A word may be defined as a portion of a string separated by one or more spaces or tabs. There will be no blank line in the input. A line may contain at most 20,000 characters. Input is terminated by EOF.

The Output

Produce a line of text for each line of input, after removing any repeated pattern. The words must be separated by only one space.

Sample Input

test string test string test string repeat repeat

Sample Output

test string repeat

很困，在於理解目容上。
首先要找到他要 reduce 的部分，按照序去找可以短的部分。
假 substring pa, pb, A

如果存在 pa A A pb 成 pa A pb。


#include <stdio.h>
#include <string.h>
#define maxN 20005
char line[maxN], *token[maxN];
int main() {
 while(gets(line)) {
 int n = 0, i, j, k;
 for(i = 0; line[i]; i++) {
 if(line[i] > 32) {
 token[n++] = line+i;
 while(line[i] > 32)
 i++;
 if(line[i] == '\0')
 break;
 line[i] = '\0';
 }
 }
 while(1) {
 int flag = 0;
 for(i = 0; i < n; i++) {
 for(j = i+1; j < n; j++) {
 int len = j-i; // [i...j-1],[j...j+len-1]
 if(j+len-1 >= n) break;
 for(k = 0; k < len; k++)
 if(strcmp(token[i+k], token[j+k]))
 break;
 if(k == len) {
 for(k = j+len, j = i+len; k < n; k++, j++)
 token[j] = token[k];
 flag = 1, i = j = n;
 n -= len;
 }
 }
 }
 if(!flag) break;
 }
 for(i = 0; i < n; i++) {
 if(i) putchar(' ');
 printf("%s", token[i]);
 }
 puts("");
 }
 return 0;
}