后缀数组 - hdu5008 Boring String Problem

题目：

http://acm.hdu.edu.cn/showproblem.php?pid=5008

题意：

将一个字符串的所有不同子串按字典序排序，多次询问第k小的子串所在位置，多个位置时输出最前的位置

思路：

先考虑如果只询问第k小的子串是什么，可以用后缀数组轻易解决，因为对于后缀i，其能产生的子串数量为len - height[i] - sa[i]，因此，用b数组存储对应sa后缀序列的每一个i值，[0,i]一共产生了多少不同子串，二分搜索第k个子串位置pos，即可得到第k个子串是什么；

然后写到这里我就被卡住了= =因为输出最前的位置直观想法就是沿着sa序列向下遍历，一直找到LCP小于sublen的为止，这样的算法在比较坑爹的数据前，比如10^5个a，时间复杂度是O(n^2)，虽然实际数据非常弱居然让这种算法过了...

正确解法应为，利用height数组特性，RMQ + 二分枚举找到下限区间r，满足在[pos,r]这段区间内的任意两个后缀的LCP大于等于sublen，再用RMQ处理sa数组，找到sa序列中[pos,r]这段区间的最小值，这种解法即使在极端数据时的时间复杂度依然只有O(nlgn)

这个算法依次用了二分求pos位置，二分+RMQ求下限r，RMQ求sa区间最小值，非常优美的思路，值得好好理解一下

代码：

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;const int MAXSIZE = 1e5 + 100;//待处理字符串，sa，rank，height均为[0,len)#define rep(i,n) for(int i = 0; i < n; i++)int rk[MAXSIZE], sa[MAXSIZE], height[MAXSIZE], wa[MAXSIZE], res[MAXSIZE];char w[MAXSIZE];  //转储待处理字符串int len;void getSa(int up) {int *k = rk, *id = height, *r = res, *cnt = wa;rep(i, up) cnt[i] = 0;rep(i, len) cnt[k[i] = w[i]]++;rep(i, up) cnt[i + 1] += cnt[i];for (int i = len - 1; i >= 0; i--) {sa[--cnt[k[i]]] = i;}int d = 1, p = 0;while (p < len){for (int i = len - d; i < len; i++) id[p++] = i;rep(i, len)  if (sa[i] >= d) id[p++] = sa[i] - d;rep(i, len) r[i] = k[id[i]];rep(i, up) cnt[i] = 0;rep(i, len) cnt[r[i]]++;rep(i, up) cnt[i + 1] += cnt[i];for (int i = len - 1; i >= 0; i--) {sa[--cnt[r[i]]] = id[i];}swap(k, r);p = 0;k[sa[0]] = p++;rep(i, len - 1) {if (sa[i] + d < len && sa[i + 1] + d < len && r[sa[i]] == r[sa[i + 1]] && r[sa[i] + d] == r[sa[i + 1] + d])k[sa[i + 1]] = p - 1;else k[sa[i + 1]] = p++;}if (p >= len) return;d <<= 1, up = p, p = 0;}}//计算rank及height值void getHeight() {int i, k, h = 0;rep(i, len) rk[sa[i]] = i;rep(i, len) {if (rk[i] == 0)h = 0;else {k = sa[rk[i] - 1];if (h) h--;while (w[i + h] == w[k + h]) h++;}height[rk[i]] = h;}}void getSuffix() {len = strlen(w);int up = 0;rep(i, len) {        w[i] = w[i] - 'a' + 1;        up = up > w[i] ? up : w[i];}w[len] = 0;getSa(up + 1);getHeight();}//nlogn时间预处理 logn时间查询区间极大极小值int ddmin[MAXSIZE][32];void RMQ_init(int A[], int len){//len 数组长度for (int i = 0; i<len; ++i){ddmin[i][0] = A[i];}for (int j = 1; (1 << j) <= len; ++j)for (int i = 0; i + (1 << j) - 1<len; ++i){ddmin[i][j] = min(ddmin[i][j - 1], ddmin[i + (1 << (j - 1))][j - 1]);}return;}int RMQ_min(int L, int R){int k = 0;while (1 << (k + 1) <= R - L + 1) k++;return min(ddmin[L][k], ddmin[R - (1 << k) + 1][k]);}//nlogn时间预处理 logn时间查询区间极大极小值int dmin[MAXSIZE][32];void LCP_init(int A[], int len){//len 数组长度for (int i = 0; i<len; ++i){dmin[i][0] = A[i];}for (int j = 1; (1 << j) <= len; ++j)for (int i = 0; i + (1 << j) - 1<len; ++i){dmin[i][j] = min(dmin[i][j - 1], dmin[i + (1 << (j - 1))][j - 1]);}return;}int LCP(int L, int R){    if (L == R) return len - sa[L];    //int l = rk[L], r = rk[R];    if (L>R) swap(L,R);    L++;int k = 0;while (1 << (k + 1) <= R - L + 1) k++;return min(dmin[L][k], dmin[R - (1 << k) + 1][k]);}long long l,r;void init(){    l = 0;    r = 0;}long long b[MAXSIZE];void calb(){    b[0] = len - sa[0];    for (int i = 1; i < len; ++i){        b[i] = b[i-1] + len - sa[i] - height[i];    }}int main(){    long long v;    int q;    while (scanf("%s",w)!=EOF){        init();        getSuffix();        calb();        LCP_init(height,len);        RMQ_init(sa,len);        scanf("%d",&q);        for (int i=0;i<q;++i){            scanf("%lld",&v);            long long k = (l^r^v)+1;            if (k>b[len-1]){                l = 0;                r = 0;                printf("0 0\n");                continue;            }            //cout<<"k: "<<k<<endl;            int L = 0, R = len-1;            int pos;            while (L<=R){                int mid = (L+R)>>1;                if (b[mid]>=k){                    pos = mid;                    R = mid -1;                }                else L = mid + 1;            }            //cout<<"pos: "<<pos<<endl;            int sublen;            if (pos == 0) sublen = k;            else sublen = height[pos] + k - b[pos-1];            //cout<<"sublen: "<<sublen<<endl;            L = pos, R = len -1;            int ll = pos, rr;            while (L<=R){                int mid = (L+R)>>1;                if (LCP(pos,mid)>=sublen){                    rr = mid;                    L = mid + 1;                }                else R = mid - 1;            }            //cout<<"ll: "<<ll<<"rr: "<<rr<<endl;            l = RMQ_min(ll,rr) + 1;            r = l + sublen - 1;            printf("%lld %lld\n",l,r);        }    }    return 0;}