[POJ3096]Surprising Strings

最新推荐文章于 2019-11-07 16:46:32 发布

转载最新推荐文章于 2019-11-07 16:46:32 发布 · 72 阅读

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原文链接：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6105568.html

本文详细解析了POJ3096 SurprisingStrings问题的算法实现，通过枚举字符串长度和字母对，并利用哈希技术判断字符串是否为Surprising类型。文章提供了完整的C++代码示例。

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[POJ3096]Surprising Strings

试题描述

The D-pairs of a string of letters are the ordered pairs of letters that are distance D from each other. A string is D-unique if all of its D-pairs are different. A string is surprising if it is D-unique for every possible distance D.

Consider the string ZGBG. Its 0-pairs are ZG, GB, and BG. Since these three pairs are all different, ZGBG is 0-unique. Similarly, the 1-pairs of ZGBG are ZB and GG, and since these two pairs are different, ZGBG is 1-unique. Finally, the only 2-pair of ZGBG is ZG, so ZGBG is 2-unique. Thus ZGBG is surprising. (Note that the fact that ZG is both a 0-pair and a 2-pair of ZGBG is irrelevant, because 0 and 2 are different distances.)

Acknowledgement: This problem is inspired by the "Puzzling Adventures" column in the December 2003 issue of Scientific American.

输入

The input consists of one or more nonempty strings of at most 79 uppercase letters, each string on a line by itself, followed by a line containing only an asterisk that signals the end of the input.

输出

For each string of letters, output whether or not it is surprising using the exact output format shown below.

输入示例

ZGBG
X
EE
AAB
AABA
AABB
BCBABCC
*

输出示例

ZGBG is surprising.
X is surprising.
EE is surprising.
AAB is surprising.
AABA is surprising.
AABB is NOT surprising.
BCBABCC is NOT surprising.

数据规模及约定

见“输入”

题解

n² 枚举一下长度和字母对，然后把字母对哈希一下，看是否出现重复。打个时间戳可以节省时间。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;

#define maxn 85
#define maxm 800
char S[maxn];
int has[maxm];

int main() {
	while(scanf("%s", S + 1) == 1) {
		int n = strlen(S + 1);
		if(n == 1 && S[1] == '*') break;
		bool ok = 1;
		memset(has, 0, sizeof(has));
		for(int l = 1; l < n; l++) {
			for(int i = 1; i + l <= n; i++) {
				int x = (S[i] - 'A') * 26 + S[i+l] - 'A';
				if(has[x] == l){ ok = 0; break; }
				has[x] = l;
			}
			if(!ok) break;
		}
		printf("%s%s\n", S + 1, ok ? " is surprising." : " is NOT surprising.");
	}
	
	return 0;
}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/6105568.html