poj 2758(后缀数组)

题意：给出一个长度为n的字母组成的序列，然后有两种操作，Q a b 询问以第a个字符为开头的后缀字符串和以第b个字符为开头的后缀字符串的lcp(最长公共前缀)的长度，I a b 表示在第b个字符之前插入字符a。
注意询问时的位置a和b是对应初始串，而比对的后缀串是当前串。
题解：思路想到了，但实现不出来 %>_<%，如果不考虑I操作，先用后缀数组处理出sa和rank，然后得到height相邻名次串的公共前缀长，然后找rank[l]+1到rank[r]的height中找最小值就是解。加上I操作就比较麻烦了，要找到l和r后面的且与l和r的距离最近的添加字符，比较这个最近距离和这两个串的lcp值：如果lcp值更小，说明添加的字符不影响他们的lcp，否则就修改l和r的值，同理递归寻找下一个距离最近的添加字符。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 50500;struct Ins {    int v, p;}ins[205];int wa[N], wb[N], ws[N], wv[N], sa[N];int rank[N], height[N], s[N], q, cnt, f[N][35];char str[N];int cmp(int* r, int a, int b, int l) {    return (r[a] == r[b]) && (r[a + l] == r[b + l]);}void DA(int *r, int *sa, int n, int m) {    int i, j, p, *x = wa, *y = wb, *t;    for (i = 0; i < m; i++) ws[i] = 0;    for (i = 0; i < n; i++) ws[x[i] = r[i]]++;    for (i = 1; i < m; i++) ws[i] += ws[i - 1];    for (i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;    for (j = 1, p = 1; p < n; j *= 2, m = p) {        for (p = 0, i = n - j; i < n; i++) y[p++] = i;        for (i = 0; i < n; i++) if (sa[i] >= j) y[p++] = sa[i] - j;        for (i = 0; i < n; i++) wv[i] = x[y[i]];        for (i = 0; i < m; i++) ws[i] = 0;        for (i = 0; i < n; i++) ws[wv[i]]++;        for (i = 0; i < m; i++) ws[i] += ws[i - 1];        for (i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i];        for (t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i++)            x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p++;    }}void calheight(int *r, int *sa, int n) {    int i, j, k = 0;    for (i = 1; i <= n; i++) rank[sa[i]] = i;    for (i = 0; i < n; height[rank[i++]] = k)        for (k ? k-- : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);}void RMQ_init(int cnt) {    for (int i = 0; i < cnt; i++)        f[i][0] = height[i];    for (int j = 1; (1 << j) <= cnt; j++)        for (int i = 0; i + (1 << j) - 1 < cnt; i++)            f[i][j] = min(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);}int RMQ(int L, int R) {    int k = 0;    while (1 << (k + 1) <= R - L + 1) k++;    return min(f[L][k], f[R - (1 << k) + 1][k]);}void Insert(int val, int pos, int len) {    int i;    for (i = 0; i < cnt; i++)        if (ins[i].p >= pos) break;        else pos--;    if (pos >= len) pos = len - 1;    for (int j = cnt; j > i; j--) ins[j] = ins[j - 1];    ins[i].v = val, ins[i].p = pos;    cnt++;}int query(int l, int r, int len) {    int x, y, res = 0;    for (x = 0; ins[x].p <= l; x++);    for (y = 0; ins[y].p <= r; y++);    if (l == r) return (len - l - 1) + (cnt - x - 1);    while (1) {        int pos1 = rank[l], pos2 = rank[r];        if (pos1 > pos2) swap(pos1, pos2);        int temp = RMQ(pos1 + 1, pos2);        int dis1 = ins[x].p - l;        int dis2 = ins[y].p - r;        int mindis = min(temp, min(dis1, dis2));        res += mindis, l += mindis, r += mindis;        if (mindis == dis1 || mindis == dis2) {            while (ins[x].p == l && ins[y].p == r) {                if (ins[x].v == ins[y].v)                    x++, y++, res++;                else return res;            }            while (ins[x].p == l) {                if (ins[x].v == s[r])                    x++, r++, res++;                else return res;            }            while (ins[y].p == r) {                if (ins[y].v == s[l])                    l++, y++, res++;                else return res;            }        }        else return res;    }    return res;}int main() {    while (scanf("%s", str) == 1) {        int len = strlen(str), maxx = -1;        for (int i = 0; i < len; i++) {            s[i] = str[i] - 'A' + 1;            maxx = max(maxx, s[i]);        }        s[len] = 0;        DA(s, sa, len + 1, maxx + 10);        calheight(s, sa, len);        RMQ_init(len + 1);        cnt = 0;        ins[cnt].v = 0;        ins[cnt++].p = N;        scanf("%d", &q);        char op[5];        int l, r;        while (q--) {            scanf("%s", op);                if (op[0] == 'Q') {                scanf("%d%d", &l, &r);                printf("%d\n", query(l - 1, r - 1, len + 1));            }            else {                scanf("%s%d", op, &l);                Insert(op[0] - 'A' + 1, l - 1, len + 1);            }        }    }    return 0;}