UVALive 4513 (LA 4513) Stammering Aliens 后缀数组 或 hash

题目大意：

白书例题

给出一个整数m>=1和一个字符串(m <= 长度 <= 40000), 找出最长的至少重复出现了m次的子串起始点和长度,可以有重叠部分地出现

如果有多个子串满足条件输出出现位置最右的

大致思路：

首先很容易想到后缀数组的做法. 利用height数组分组, 很简单就不说了, 细节见代码

另外白书上说可以用hash, 所以用这个题试了一下hash

表示第一次用hash...然后就坑了有木有= =

刚开始我取得是H函数中x = 2的情况来写

这样可以用位运算..然后交了好多次之后发现取x = 2就是一个错误...

后来把白书中的x从123改成2发现也WA了, 177,233什么的都可以过= =.....于是第一次写hash就被概率性地挂掉了...

两种方法的代码如下：

后缀数组解法:

Result  :  Accepted     Memory  :  ? KB     Time  :  193 ms

/*
 * Author: Gatevin
 * Created Time:  2015/2/11 19:04:46
 * File Name: Mononobe_Mitsuki.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 maxn 42333

/*
 * Doubling Algorithm求后缀数组
 */
int 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 *= 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 = 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;
}
    
int s[maxn], sa[maxn];
char in[maxn];
int pos[maxn];
int M;

bool check(int mid, int n, int& start)//对于连续出现M - 1次以上height值>=mid的情况更新start的答案
{
    int cnt = 1;
    bool ret = false;
    int tmp = 0;
    for(int i = 1; i <= n; i++)
        if(height[i] >= mid)
        {
            tmp = max(tmp, sa[i]);//当前连续片中的最右串
            cnt++;
            if(cnt >= M)
            {
                ret = true;
                start = max(start, tmp);//更新start位置
            }
        }
        else
            cnt = 1, tmp = sa[i];
    return ret;
}


int main()
{
    while(scanf("%d", &M), M)
    {
        scanf("%s", in);
        int n = strlen(in);
        if(M == 1)
        {
            printf("%d 0\n", n);
            continue;
        }
        for(int i = 0; i < n; i++)
            s[i] = in[i] - 'a' + 1;
        s[n] = 0;
        da(s, sa, n + 1, 27);
        calheight(s, sa, n);
        int L = 1, R = n, len = 0, mid;//二分判断是否存在长度为mid的字符串出现了M次以上
        while(L <= R)
        {
            mid = (L + R) >> 1;
            pos[mid] = 0;
            if(check(mid, n, pos[mid]))
            {
                len = mid;
                L = mid + 1;
            }
            else
                R = mid - 1;
        }
        if(len == 0)
            printf("none\n");
        else
            printf("%d %d\n", len, pos[len]);
    }
    return 0;
}

hash的做法:

Result  :  Accepted     Memory  :  ? KB     Time  :  1319 ms

/*
 * Author: Gatevin
 * Created Time:  2015/2/11 19:46:52
 * File Name: Mononobe_Mitsuki.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;
typedef unsigned long long ulint;
#define maxn 42333

const ulint x = 233;

/*
 * 刚开始取得x = 2然后就各种悲剧= =...
 * 因为当时想的x = 2的时候用位运算可能快一些
 * 同时也利用了unsigned long long的溢出取模
 * 然后跪了好几下实在没救了= =
 * 真是醉了...
 */
int M;
char in[maxn];
ulint hash[maxn];
int pos[maxn];
int id[maxn];
ulint xp[maxn];
ulint H[maxn];

bool cmp(int a, int b)
{
    return hash[a] < hash[b];
}

bool check(int mid, int n)
{
    pos[mid] = 0;
    id[0] = 0;
    for(int i = 0; i + mid - 1 < n; i++)//就算每一段的hash值
        id[i] = i, hash[i] = H[i] - H[i + mid]*xp[mid];
    sort(id, id + n - mid + 1, cmp);//按hash值排序
    int cnt = 1;
    bool ret = false;
    int tmp = id[0];
    for(int i = 1; i <= n - mid; i++)
        if(hash[id[i]] == hash[id[i - 1]])
        {
            cnt++;
            tmp = max(tmp, id[i]);//注意片段性的最大tmp中取最大的作为pos[mid]
            if(cnt >= M)
                pos[mid] = max(pos[mid], tmp), ret = true;
        }
        else
            cnt = 1, tmp = id[i];
    return ret;
}


int main()
{
    xp[0] = 1;
    for(int i = 1; i <= 40000; i++)//预处理x次幂项
        xp[i] = xp[i - 1]*x;
    while(scanf("%d", &M), M)
    {
        scanf("%s", in);
        int n = strlen(in);
        if(M == 1)
        {
            printf("%d 0\n", n);
            continue;
        }
        H[n] = 0;
        for(int i = n - 1; i >= 0; i--)//预处理H[i] = s[i] + s[i + 1]*x + ... + s[n]*x^(n - i)
            H[i] = H[i + 1]*x + (in[i] - 'a');
        
        int L = 1, R = n, mid, len = 0;
        while(L <= R)
        {
            mid = (L + R) >> 1;
            if(check(mid, n))
            {
                len = mid;
                L = mid + 1;
            }
            else
                R = mid - 1;
        }
        if(len == 0)
            printf("none\n");
        else
            printf("%d %d\n", len, pos[len]);
    }
    return 0;
}