UVA - 12338 Anti-Rhyme Pairs （哈希）

Description

D

Anti-Rhyme Pairs Input: Standard Input Output: Standard Output

 

Often two words that rhyme also end in the same sequence of characters. We use this property to define the concept of an anti-rhyme. An anti-rhyme is a pair of words that have a similar beginning. The degree of anti-rhyme of a pair of words is further defined to be the length of the longest string S such that both strings start withS. Thus, “arboreal” and “arcturus” are an anti-rhyme pair of degree 2, while “chalkboard” and “overboard” are an anti-rhyme pair of degree 0.

 

You are given a list of words. Your task is, given a list of queries in the form(i, j), print the degree of anti-rhyme for the pair of strings formed by thei-th and the j-th words from the list.

 

Input

Input consists of a number of test cases. The first line of input contains the number of test casesT (T ≤ 35). Immediately following this line are T cases.

 

Each case starts with the number of strings N (1 ≤ N ≤ 10⁵) on a line by itself. The followingN lines each contain a single non-empty string made up entirely of lower case English characters ('a' to 'z'), whose lengthL is guaranteed to be less than or equal to 10,000. In every case it is guaranteed thatN*L ≤ 10⁶.

 

The line following the last string contains a single integer Q (1 ≤ Q ≤ 10⁶), the number of queries. Each of theQ lines following contain a query made up of two integers i and j separated by whitespace (1 ≤ i, j ≤ N).

 

Output

The output consists of T cases, each starting with a single line with“Case X:”, where X indicates the X-th case. There should be exactlyQ lines after that for each case. Each of those Q lines should contain an integer that is the answer to the corresponding query in the input.

 

Sample Input                            Output for Sample Input

2 5 daffodilpacm daffodiliupc distancevector distancefinder distinctsubsequence 4 1 2 1 5 3 4 4 5 2 acm icpc 2 1 2 2 2

Case 1: 8 1 8 4 Case 2: 0 4

题意：给你n个字符串，让你求给定的两个串的最长公共前缀
思路：预处理是无法达到时间要求的，那么我们能够先hash处理出每一个字符串每一个字符的哈希值（这个值是递增），然后就能二分的比較查询了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long ll;
using namespace std;
const int maxn = 1000005;
const int seed = 31;

char str[maxn];
int start[maxn], len[maxn];
ll hash[maxn];

int cal(int a, int b) {
	int l = 0, r = min(len[a], len[b]);	
	while (l < r) {
		int m = (l + r + 1) >> 1;
		if (hash[start[a]+m-1] == hash[start[b]+m-1]) 
			l = m;
		else r = m-1;
	}
	return l;
}

int main() {
	int t, n, cas = 1;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);	
		int cur = 0;
		for (int i = 0; i < n; i++) {
			scanf("%s", str+cur);
			len[i] = strlen(str+cur);
			start[i] = cur;
			hash[cur] = str[cur] - 'a';
			for (int j = 1; j < len[i]; j++)
				hash[cur+j] = hash[cur+j-1] * seed + str[cur+j] - 'a';
			cur += len[i];
		}
		scanf("%d", &n);
		printf("Case %d:\n", cas++);
		int a, b;
		while (n--) {
			scanf("%d%d", &a, &b);	
			printf("%d\n", cal(a-1, b-1));
		}
	}
	return 0;
}