（后缀数组）HDU 6194-string string string

string string string

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 774    Accepted Submission(s): 225

Problem Description

Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.

 

Input

The first line contains an integer T (T≤100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.

 

Output

For each test case, print the number of the important substrings in a line.

 

Sample Input

2 2 abcabc 3 abcabcabcabc

 

Sample Output

6 9

只需求出出现次数>=k和 >=k+1的个数。

对于求出现次数>=x的，如果x=1，则直接特判输出。

若不然，只需维护长度x-1的height序列（height序列长度x-1 则对应的后缀数即为x），维护序列中height的最小值，按下标从小到大扫，注意到每次更新，只有该最小值发生变大时，才会增加新的出现次数>=x次的子串。（不变或变小时当前的子串之前都计数过）

以此计算>=x >=x+1的相减即可。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <bitset>
#include <utility>
#include <assert.h>
using namespace std;
#define rank rankk
#define mp make_pair
#define pb push_back
#define xo(a,b) ((b)&1?(a):0)
#define tm tmp
//#define LL ll
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef long long ll;
typedef pair<ll,int> pli;
typedef pair<ll,ll> pll;
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3fll;
const int MAX=1e5+5;
//const ll MAXN=2e8;
const int MAX_N=MAX;
const ll MOD=998244353;
//const long double pi=acos(-1.0);
//const double eps=0.00000001;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
template<typename T>inline T abs(T a) {return a>0?a:-a;}
template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
inline ll powMM(ll a,ll b,ll M){
    ll ret=1;
    a%=M;
//    b%=M;
    while (b){
        if (b&1) ret=ret*a%M;
        b>>=1;
        a=a*a%M;
    }
    return ret;
}
void open()
{
//    freopen("1009.in","r",stdin);
    freopen("out.txt","w",stdout);
}
const int MAXN=100005;
int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
//除s[n-1]外的所有s[i]都大于0，r[n-1]=0
//函数结束以后结果放在sa数组中
bool cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int str[],int sa[],int rank[],int height[],int n,int m)
{
    str[n++]=0;
    int i, j, p, *x = t1, *y = t2;
    //第一轮基数排序，如果s的最大值很大，可改为快速排序
    for(i = 0;i < m;i++)c[i] = 0;
    for(i = 0;i < n;i++)c[x[i] = str[i]]++;
    for(i = 1;i < m;i++)c[i] += c[i-1];
    for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;
    for(j = 1;j <= n; j <<= 1)
    {
        p = 0;
        //直接利用sa数组排序第二关键字
        for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
        for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
        //这样数组y保存的就是按照第二关键字排序的结果
        //基数排序第一关键字
        for(i = 0; i < m; i++)c[i] = 0;
        for(i = 0; i < n; i++)c[x[y[i]]]++;
        for(i = 1; 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);
        //根据sa和x数组计算新的x数组 swap(x,y);
        p = 1; x[sa[0]] = 0;
        for(i = 1;i < n;i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
        if(p >= n)break;
        m = p;//下次基数排序的最大值
    }
    int k = 0; n--;
    for(i = 0;i <= n;i++)rank[sa[i]] = i;
    for(i = 0;i < n;i++)
    {
        if(k)k--;
        j = sa[rank[i]-1]; while(str[i+k] == str[j+k])k++; height[rank[i]] = k;
    }
}
int rank[MAXN],height[MAXN];
int RMQ[MAXN];
int mm[MAXN];
int best[20][MAXN];
void initRMQ(int n)
{
    mm[0]=-1;
    for(int i=1;i<=n;i++)
    mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
    for(int i=1;i<=n;i++)best[0][i]=i;
    for(int i=1;i<=mm[n];i++)
    for(int j=1;j+(1<<i)-1<=n;j++)
    {
        int a=best[i-1][j];
        int b=best[i-1][j+(1<<(i-1))];
        if(RMQ[a]<RMQ[b])best[i][j]=a; else best[i][j]=b;
    }
}
int askRMQ(int a,int b)
{
    int t; t=mm[b-a+1];
    b-=(1<<t)-1;
    a=best[t][a];b=best[t][b];
    return RMQ[a]<RMQ[b]?a:b;
}
int lcp(int a,int b)
{
    a=rank[a];b=rank[b];
    if(a>b)swap(a,b);
    return height[askRMQ(a+1,b)];
}
int t,k,len;
char sts[MAXN];
int st[MAXN],sa[MAXN];
multiset<int>s;
ll cal(int x)
{
    ll an=0;
    if(x==1)
    {
        an=1LL*len*(len+1LL)/2LL;
        for(int i=1;i<=len;i++)an-=height[i];
    }
    else
    {
        s.clear();
        for(int i=1;i<=x-1;i++)s.insert(height[i]);
        for(int i=x;i<=len;i++)
        {
            int pre=*s.begin();s.erase(s.find(height[i-x+1]));s.insert(height[i]);
            printf("%d %d!
",pre,*s.begin());
            an+=max(0,*s.begin()-pre);
        }
    }
    return an;
}
int main()
{
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&k);
        scanf("%s",sts);len=strlen(sts);
        memset(height,0,sizeof(height));
        for(int i=0;i<len;i++)st[i]=sts[i]-'a'+1;
        da(st,sa,rank,height,len,27);
        for(int i=1;i<=len;i++)
            printf("%d
",height[i]);
        printf("%lld
",cal(k)-cal(k+1));
    }
}