hdu 5008 Boring String Problem(后缀数组)

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题目链接：hdu 5008 Boring String Problem

题目大意：给定一个字符串，初始状态l，r为0，每次询问子串中字典序第l^r^v+1的子串区间，对于重复的输出下标小的。

解题思路：后缀数组，对给定字符串做后缀数组，然后根据height数组确定每个位置做为起点的子串有多少，然后二分查找确定起点位置，但是因为子串的重复的要输出下表小的，所以确定起点后还要确定字典序最小的下标。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long ll;const int maxn = 1e5+5;struct Suffix_Arr {    int n, s[maxn];    int SA[maxn], rank[maxn], height[maxn];    int tmp_one[maxn], tmp_two[maxn], c[maxn];    ll sum, num[maxn];    void init (char* str);    int idx(char ch);    void build(int m);    void get_height();    void get_substring();    void query(ll x, ll& L, ll& R);}AC;int n;ll l, r, x;char str[maxn];int main () {    while (scanf("%s", str) == 1) {        AC.init(str);        AC.build(27);        AC.get_height();        AC.get_substring();        scanf("%d", &n);        l = r = 0;        while (n--) {            scanf("%I64d", &x);            AC.query((x^l^r) + 1, l, r);            printf("%I64d %I64d\n", l, r);        }    }    return 0;}void Suffix_Arr::query(ll x, ll& L, ll& R) {    if (x >= sum) {        L = R = 0;        return;    }    int l = 0, r = n;    for (int i = 0; i < 100; i++) {        int mid = (l + r) / 2;        if (num[mid] < x)            l = mid;        else            r = mid;    }    int len = height[l] + x - num[l], pos = SA[l];    for (int i = l + 1; height[i] >= len && i < n; i++)        pos = min(SA[i], pos);    for (int i = l - 1; height[i+1] >= len && i >= 0; i--)        pos = min(SA[i], pos);    L = pos + 1;    R = pos + len;}void Suffix_Arr::get_substring() {    num[0] = 0;    for (int i = 1; i <= n; i++)        num[i] = num[i-1] + (n - 1 - SA[i-1]) - height[i-1];    sum = num[n];}void Suffix_Arr::init (char* str) {    n = strlen(str);    for (int i = 0; i < n; i++)        s[i] = idx(str[i]);    s[n++] = 0;}int Suffix_Arr::idx(char ch) {    return ch - 'a' + 1;}void Suffix_Arr::get_height() {    for (int i = 0; i < n; i++)        rank[SA[i]] = i;    int mv = 0;    for (int i = 0; i < n; i++) {        if (mv) mv--;        if (rank[i] == 0) {            height[0] = 0;            continue;        }        int j = SA[rank[i]-1];        while (s[i+mv] == s[j+mv])            mv++;        height[rank[i]] = mv;    }}void Suffix_Arr::build (int m) {    int *x = tmp_one, *y = tmp_two;    for (int i = 0; i < m; i++) c[i] = 0;    for (int i = 0; i < n; i++) c[x[i] = s[i]]++;    for (int i = 1; i < m; i++) c[i] += c[i-1];    for (int i = n - 1; i >= 0; i--) SA[--c[x[i]]] = i;    for (int k = 1; k <= n; k <<= 1) {        int mv = 0;        for (int i = n - k; i < n; i++) y[mv++] = i;        for (int i = 0; i < n; i++) if (SA[i] >= k)            y[mv++] = SA[i] - k;        for (int i = 0; i < m; i++) c[i] = 0;        for (int i = 0; i < n; i++) c[x[y[i]]]++;        for (int i = 1; i < m; i++) c[i] += c[i-1];        for (int i = n - 1; i >= 0; i--) SA[--c[x[y[i]]]] = y[i];        swap(x, y);        mv = 1;        x[SA[0]] = 0;        for (int i = 1; i < n; i++)            x[SA[i]] = (y[SA[i-1]] == y[SA[i]] && y[SA[i-1] + k] == y[SA[i] + k] ? mv - 1 : mv++);        if (mv >= n)            break;        m = mv;    }}

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