UVA-644 Immediate Decodability

最新推荐文章于 2021-07-28 19:02:08 发布

原创最新推荐文章于 2021-07-28 19:02:08 发布 · 239 阅读

0 ·

CC 4.0 BY-SA版权

acm 专栏收录该内容

25 篇文章

订阅专栏

本文介绍了解决UVA-537问题的方法，该问题是关于判断一组二进制字符串是否立即可解码。通过对比字符串，确定它们之间是否存在前缀关系来解决这个问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

UVA-537 Artificial Intelligence?

题目大意：如果输入的单词中，有的单词是其它单词的前缀，则 is not immediately decodable ，反之则是 is 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

解题思路：逐个比较即可

//UVA-644 Immediate Decodability
#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std;

char m[10005][10005];
char n[10005];
int count = 1;

int main() {
    while (scanf ("%s",n) != EOF) {
        int t = 0;
        while(strcmp(n,"9\0") != 0) {
            strcpy(m[t++], n);
            scanf("%s",n);
        }
        int flag = 1;
        for (int i = 0; i < t; i++)
            for (int j = i + 1; j < t; j++) {
                int len1 = strlen (m[i]);
                int len2 = strlen (m[j]);
                int k;
                for (k = 0; k < min (len1,len2); k++) {
                    if (m[i][k] != m[j][k])
                        break;
                    if (k == min(len1,len2) - 1) {
                        flag = 0;
                    }
                }
            }
        if (flag)
            printf ("Set %d is immediately decodable\n",count++);
        else
            printf ("Set %d is not immediately decodable\n",count++);
    }
    return 0;
}