每次询问求出两个字符串的最长公共前缀的长度 后缀数组+RMQ+lcp UVA 12338 - Anti-Rhyme Pairs

题目链接

题意：给定一些字符串，每次询问求出两个字符串的最长公共前缀的长度

思路：把字符串排序，就能求出height和rank数组，然后利用RMQ查询即可

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

const int N = 100005;

typedef pair<string, int> pii;

pii str
;
int save
;

int t, n, height
, rank
;

void init() {
scanf("%d", &n);
for (int i = 0; i < n; i++) {
cin >> str[i].first;
save[i] = str[i].first.length();
str[i].second = i;
}
sort(str, str + n);
for (int i = 0; i < n; i++) {
rank[str[i].second] = i;
if (i == 0) continue;
int len = min(str[i - 1].first.length(), str[i].first.length());
int j;
for (j = 0; j < len; j++) {
if (str[i - 1].first[j] != str[i].first[j])
break;
}
height[i] = j;
}
}

int best[N * 10][20];

void initRMQ() {
for (int i = 0; i < n; i++) best[i][0] = height[i];
for (int j = 1; (1<<j) <= n; j++)
for (int i = 0; i + (1<<j) - 1 < n; i++)
best[i][j] = min(best[i][j - 1], best[i + (1<<(j - 1))][j - 1]);
}

int lcp(int L, int R) {
L--; R--;
if (L == R) return save[L];
L = rank[L]; R = rank[R];
if (L > R) swap(L, R);
L++;
int k = 0;
while ((1<<(k + 1)) <= R - L + 1) k++;
return min(best[L][k], best[R - (1<<k) + 1][k]);
}

int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
init();
initRMQ();
int q, l, r;
scanf("%d", &q);
printf("Case %d:\n", ++cas);
while (q--) {
scanf("%d%d", &l, &r);
printf("%d\n", lcp(l, r));
}
}
return 0;
}