HDU 5008 Boring String Problem 后缀数组

题目大意：

就是给出一个字符串(长度不超过10W), 接着Q次询问, 对于每次询问要找到第K小子串, 注意这里的字串是本质不同的子串, 

大致思路：

由于Q可以达到10^5 使用后缀自动机的话容易超时, 于是需要使用后缀数组, 对于给出的字符串求出后缀数组之后利用height数组处理出前i个分支下有多少个不同的子串, 那么对于每次询问二分查找即可, 由于每次回答的[l, r]中l药最小, 二分之后向后再找一点距离即可

代码如下：

Result  :  Accepted     Memory  :  5488 KB     Time  :  592 ms

/* * Author: Gatevin * Created Time:  2015/5/4 15:22:39 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl#define rank rankrankrank#define maxn 100010int wa[maxn], wb[maxn], wv[maxn], Ws[maxn];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 *x = wa, *y = wb, *t, i, j, p;    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 <<= 1, 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 = 1; 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++;    }    return;}int rank[maxn], height[maxn];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++);    return;}char in[maxn];int s[maxn], sa[maxn];lint cnt[maxn];lint v, l, r, k;void init(int len){    cnt[0] = 0;    for(int i = 1; i <= len; i++)        cnt[i] = cnt[i - 1] + (len - sa[i]) - height[i];    return;}void solve(int len)//找出第k大的位置{    if(k > cnt[len])    {        l = r = 0;        return;    }    int L = 0, R = len, mid, res;    while(L <= R)    {        mid = (L + R) >> 1;        if(cnt[mid] < k)        {            res = mid;            L = mid + 1;        }        else R = mid - 1;    }    int ad = k - cnt[res];    int LEN = ad + height[res + 1];//这个子串的长度    l = sa[res + 1];    res += 2;    while(height[res] >= LEN)//向后找使得l最小        l = min(l, (lint)sa[res]), res++;    r = l + LEN - 1;    l++, r++;}int main(){    while(scanf("%s", in) != EOF)    {        int len = strlen(in);        for(int i = 0; i < len; i++)            s[i] = in[i] - 'a' + 1;        s[len] = 0;        da(s, sa, len + 1, 28);        calheight(s, sa, len);        init(len);        int Q;        scanf("%d", &Q);        l = r = 0;        while(Q--)        {            scanf("%I64d", &v);            k = (l^r^v) + 1;            solve(len);            printf("%I64d %I64d\n", l, r);        }    }    return 0;}