hdu 5853 Jong Hyok and String （二分+后缀数组+RMQ）

最新推荐文章于 2018-08-12 20:51:51 发布

原创最新推荐文章于 2018-08-12 20:51:51 发布 · 512 阅读

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本文解析了一道关于字符串匹配的编程竞赛题目，介绍了解题思路和实现细节，包括字符串倒序连接、后缀数组构建、二分查找及RMQ算法的应用。

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Jong Hyok and String

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 301 Accepted Submission(s): 66

Problem Description

Jong Hyok loves strings. One day he gives a problem to his friend you. He writes down n strings Pi in front of you, and asks m questions. For i-th question, there is a string Qi. We called strange set(s) = {(i, j) | s occurs in Pi and j is the position of its last character in the current occurence}. And for ith question, you must answer the number of different strings t which satisfies strange set(Qi) = strange set(t) and t is a substring of at least one of the given n strings.

Input

First line contains T, a number of test cases.

For each test cases, there two numbers n, m and then there are n strings Pi and m strings Qj.(i = 1…n, j = 1…m)

1 <= T <= 10
1 <= n <= 100000
1 <= m<= 500000
1 <=|Pi|<=100000
1 <=|Qi|<=100000

∑ni=1|Pi|≤100000
File size is less than 3.5 megabytes.

Output

For each test case, first line contains a line “Case #x:”, x is the number of the case.

For each question, you should print one integer in one line.

Sample Input

1
2 2
aba
ab
a
ab

Sample Output

Case #1:
1
2
Hint

strange set(“a”) ={(1, 1), (1, 3), (2, 1)}.
strange set(“ab”) ={(1, 2), (2, 2)}.
strange set(“b”) ={(1, 2), (2, 2)}.




分析：考虑到P串长度总和为10w，我们把每个P串倒着连成一串，中间用特殊符隔开，由于特殊符号需要10w个，所以我们要把字符换成int，然后跑后缀数组，就可以得到每个后缀的排名，查询时也把Q串倒序，并换成int，然后去每个后缀里面找哪些可以和这个倒序的Q串匹配，由于每个后缀已经排好序，所以我们用二分查找得到下界lower和上界upper，如果找不到，那么答案就是0。先定义一个maxlen，如果lower==upper，那么maxlen为排名为lower的这个后缀的不包含特殊符的最大前缀，如果lower<upper，那么maxlen=min（height[lower+1]...height[upper])，这里取最小值可以用RMQ处理height值得到，然后最后的答案ans=maxlen-max（height[lower],height[upper+1])
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
const int N=200010;
const int maxn=100010;
int cmp(int *r,int a,int b,int l)
{
    return (r[a]==r[b])&&(r[a+l]==r[b+l]);
}
int wa[N],wb[N],ws[N],wv[N];
int Rank[N],height[N];
void DA(int *r,int *sa,int n,int m)
{
    int i,j,p,*x=wa,*y=wb,*t;
    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++;
    }
    int k=0;
    for(i=1;i<n;i++) Rank[sa[i]]=i;
    for(i=0;i<n-1; height[Rank[i++]] = k )
    for(k?k--:0,j=sa[Rank[i]-1]; r[i+k]==r[j+k]; k++);
}
char s[maxn];
int a[maxn];
int r[N],sa[N],C[N][20];
int pos[N];
void rmq(int nn)
{
    for(int i=1;i<nn;i++) C[i][0]=height[i];
    for(int j=1;(1<<j)<nn;j++)
        for(int i=1;i+(1<<j)-1<nn;i++)
        C[i][j]=min(C[i][j-1],C[i+(1<<(j-1))][j-1]);
}
int getmin(int l,int r)
{
    int k=0;
    while((1<<(k+1))<=r-l+1) k++;
    return min(C[l][k],C[r-(1<<k)+1][k]);
}
int cmp(int pos,int len,int num)
{
    for(int i=0;i<len;i++)
    {
        if(r[i+pos]>a[i]) return 1;
        else if(r[i+pos]<a[i]) return -1;
    }
    return 0;
}
int main()
{
    int T;
    scanf("%d",&T);
    for(int ca=1;ca<=T;ca++)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int num=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",s);
            int len=strlen(s);
            reverse(s,s+len);
            int mm=num+len-1;
            for(int j=0;j<len;j++)
            {
                pos[num]=mm;
                r[num++]=s[j]-'a'+1;
            }
            r[num++]=26+i;
        }
        r[num++]=0;
        DA(r,sa,num,n+27);
        height[num]=0;
        rmq(num);
        printf("Case #%d:\n",ca);
        while(m--)
        {
            scanf("%s",s);
            int len=strlen(s);
            reverse(s,s+len);
            for(int i=0;i<len;i++)
                a[i]=s[i]-'a'+1;
            int lower=-1,upper=-1;
            int L,R;
            L=1,R=num-1;
            while(L<=R)
            {
                int mid=(L+R)>>1;
                int k=cmp(sa[mid],len,num);
                if(k==0)
                {
                    lower=mid;
                    R=mid-1;
                }
                else if(k==1) R=mid-1;
                else L=mid+1;
            }
            L=1,R=num-1;
            while(L<=R)
            {
                int mid=(L+R)>>1;
                int k=cmp(sa[mid],len,num);
                if(k==0)
                {
                    upper=mid;
                    L=mid+1;
                }
                else if(k==1) R=mid-1;
                else L=mid+1;
            }
            if(lower==-1)
            {
                puts("0");
                continue;
            }
            if(lower==upper)
            {
                printf("%d\n",(pos[sa[lower]]-sa[lower]+1)-max(height[lower],height[lower+1]));
                continue;
            }
            int ans=getmin(lower+1,upper)-max(height[lower],height[upper+1]);
            printf("%d\n",ans);
        }
    }
    return 0;
}