uva 12338 - Anti-Rhyme Pairs(后缀数组+RMQ)

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题目链接：uva 12338 - Anti-Rhyme Pairs

题目大意：给定若干个字符串，每次询问两个字符串的最长公共前缀。

解题思路：本来应该将每个字符串连接起来做后缀数组，但其实可以直接把一个字符串看成是一个字符，然后排序了就对应是SA数组，然后处理height即可。然后根据后缀数组的性质，字符串i和j的最长公共前缀长度即为rank[i]+1~rank[j]之间height的最小值。特判i=j的情况。

#include <cstdio>#include <cstring>#include <string>#include <vector>#include <iostream>#include <algorithm>using namespace std;typedef pair<string, int> pii;const int maxn = 1e5+5;int Rank[maxn];int dp[maxn][20];vector<pii> g;vector<int> height;void rmq_init(const vector<int>& A) {    int n = A.size();    for (int i = 0; i < n; i++)        dp[i][0] = A[i];    for (int j = 1; (1<<j) <= n; j++) {        for (int i = 0; i + (1<<j) - 1 < n; i++)            dp[i][j] = min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);    }}int rmq_query (int l, int r) {    if (l == r)        return g[l].first.length();    if (l > r)        swap(l, r);    l++;    int k = 0;    while ((1<<(k+1)) <= r - l + 1)        k++;    return min(dp[l][k], dp[r - (1<<k) + 1][k]);}void solve () {    sort(g.begin(), g.end());    for (int i = 0; i < g.size(); i++)        Rank[g[i].second] = i;    height.clear();    height.push_back(0);    for (int i = 1; i < g.size(); i++) {        int n = min(g[i].first.length(), g[i-1].first.length());;        int j;        for (j = 0; j < n; j++)            if (g[i].first[j] != g[i-1].first[j])                break;        height.push_back(j);    }    rmq_init(height);}int main () {    int cas, n, l, r;    string s;    cin >> cas;    for (int kcas = 1; kcas <= cas; kcas++) {        cin >> n;        g.clear();        for (int i = 0; i < n; i++) {            cin >> s;            g.push_back(make_pair(s, i));        }        solve();        cout << "Case " << kcas << ":" << endl;        cin >> n;        while (n--) {            cin >> l >> r;            cout << rmq_query(Rank[l-1], Rank[r-1]) << endl;        }    }    return 0;}

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