[hdu4622 Reincarnation]后缀数组

题意：给一个长度为2000的字符串，10000次询问区间[L,R]内的不同子串的个数

思路：对原串的每个前缀求一边后缀数组，询问[L,R]就变成了询问[L,n]了，即求一个后缀里面出现了多少个不同子串。于是对所有大于等于L的后缀统计一遍即可。

 

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#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}
template<typename T>
void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];}
template<typename T>
void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];}

const double PI = acos(-1.0);
const int INF = 1e9 + 7;

/* -------------------------------------------------------------------------------- */

const int maxn = 2e3 + 7;

struct SA {
    //const static int maxn = 2e3 + 7;
    int sa[maxn], t[maxn], t2[maxn], c[maxn], n, m;
    int Rank[maxn], Height[maxn];
    int s[maxn];

    void init(int n, int m, char s[]) {
        for (int i = 0; i < n; i ++) this->s[i] = s[i];
        this->s[n] = 0;
        this->n = n + 1;
        this->m = m;
    }

    void build() {
        int i, *x = t, *y = t2;
        for (i = 0; i < m; i ++) c[i] = 0;
        for (i = 0; i < n; i ++) c[x[i] = s[i]] ++;
        for (i = 0; i < m; i ++) c[i] += c[i - 1];
        for (i = n - 1; i >= 0; i --) sa[-- c[x[i]]] = i;
        for (int k = 1; k <= n; k <<= 1) {
            int p = 0;
            for (i = n - k; i < n; i ++) y[p ++] = i;
            for (i = 0; i < n; i ++) if (sa[i] >= k) y[p ++] = sa[i] - k;
            for (i = 0; i < m; i ++) c[i] = 0;
            for (i = 0; i < n; i ++) c[x[y[i]]] ++;
            for (i = 0; i < m; i ++) c[i] += c[i - 1];
            for (i = n - 1; i >= 0; i --) sa[-- c[x[y[i]]]] = y[i];
            swap(x, y);
            p = 1; x[sa[0]] = 0;
            for (i = 1; i < n; i ++) {
                x[sa[i]] = y[sa[i - 1]] == y[sa[i]] &&
                            y[sa[i - 1] + k] == y[sa[i] + k]? p - 1 : p ++;
            }
            if (p >= n) break;
            m = p;
        }
    }
    void getHeight() {
        int i, k = 0;
        for (i = 0; i < n; i ++) Rank[sa[i]] = i;
        for (i = 0; i < n; i ++) {
            if (k) k --;
            int j = sa[Rank[i] - 1];
            while (s[i + k] == s[j + k]) k ++;
            Height[Rank[i]] = k;
        }
    }
};
SA sa;
int H[maxn][maxn], S[maxn][maxn];
char s[maxn];

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    int T, m;
    cin >> T;
    while (T --) {
        scanf("%s", s);
        for (int i = 0; s[i]; i ++) {
            sa.init(i + 1, 128, s);
            sa.build();
            sa.getHeight();
            copy(H[i], sa.Height);
            copy(S[i], sa.sa);
        }

        cin >> m;
        while (m --) {
            int u, v;
            scanf("%d%d", &u, &v);
            int *ph = H[v - 1], *ps = S[v - 1], common = 0, ans = 0;
            u --;
            for (int i = 1; i <= v; i ++) {
                if (ps[i] >= u) {
                    ans += v - ps[i] - common;
                    common = INF;
                }
                umin(common, ph[i + 1]);
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}
posted @ 2015-08-12 14:39  jklongint  阅读(390)  评论(0)  编辑  收藏  举报