UVA 12338 Anti-Rhyme Pairs (最长公共前缀)

Anti-Rhyme Pairs

Time Limit: 5000ms

Memory Limit: 131072KB

Often two words that rhyme alsoend in the same sequence of characters. We use this property to define theconcept 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 lengthof the longest string S such thatboth strings start with S. Thus,arboreal and arcturus are an anti-rhyme pair of degree 2, whilechalkboard and overboard arean 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 stringsformed by the i-th and the j-th words from the list.

 

Input

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

Each case starts with the numberof strings N (1 ≤ N ≤10⁵) on a line by itself. The following N lines each contain a single non-emptystring made up entirely of lower case English characters ('a' to 'z'), whoselength L is guaranteed to be lessthan or equal to 10,000. In everycase it is guaranteed that N*L ≤10⁶.

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

 

Output

The outputconsists of T cases, each starting witha single line with Case X:, where X indicates the X-th case. There should be exactly Q lines after that for each case. Each of those Q lines should contain an integer thatis the answer to the corresponding query in the input.

 

SampleInput 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

Warning:I/O files is huge, make sure your I/O is fast.

题目链接：https://acm.bnu.edu.cn/v3/contest_show.php?cid=7985#problem/D

题目大意：给n个字符串，m个查询，求第x个和第y个的最长公共前缀长度

题目分析：对字符串按前缀hash，二分长度判断即可

//2018年2月28日09:21:49
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;

const int maxn = 100100;
const int seed = 131;
const int mod = 1e9+7;

typedef long long ll;
typedef unsigned long long ull;
int T, n, m;
char s[maxn];
vector<ull> v[maxn];

int hash(char *s, int id){
    int Hash = 0;
    for(int i=0; s[i]; i++){
        Hash = (Hash*seed + s[i]-'0') % mod;
        v[id].push_back(Hash); 
    }
    return Hash;
}

int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        for(int i=1;i<=n;i++){
            scanf("%s", s);
            v[i].clear();
            hash(s, i);
        }
        scanf("%d", &m);
        while(m--){
            int a, b;
            scanf("%d%d", &a, &b);
            int l=0, r=min(v[a].size(), v[b].size())-1, ans=0;
            while(l <= r){
                int mid = l+r >> 1;
                if(v[a][mid] == v[b][mid]) ans=mid, l=mid+1; 
                else r = mid-1;
            }
            if(v[a][ans] != v[b][ans])
                printf("%d
", ans);
            else printf("%d
", ans+1);
        }
    }

    return 0;
}