hdu 5790 Prefix (主席树维护区间内不同数的个数)

Prefix

Time Limit: 2000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1103    Accepted Submission(s): 326

 

Problem Description

Alice gets N strings. Now she has Q questions to ask you. For each question, she wanna know how many different prefix strings between Lth and Rth strings. It's so easy right? So solve it!

 

Input

The input contains multiple test cases.

For each test case, the first line contains one integer N(1≤N≤100000). Then next N lines contain N strings and the total length of N strings is between 1 and 100000. The next line contains one integer Q(1≤Q≤100000). We define a specail integer Z=0. For each query, you get two integer L, R(0=<L,R<N). Then the query interval [L,R] is [min((Z+L)%N,(Z+R)%N)+1,max((Z+L)%N,(Z+R)%N)+1]. And Z change to the answer of this query.

 

Output

For each question, output the answer.

 

Sample Input

3 abc aba baa 3 0 2 0 1 1 1

 

Sample Output

7 6 3

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

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思路： 因为n个字符串总长度最大为1e5 ，所以可以将每个前缀哈希一下，或者用trie 映射一下，就可以为每个前缀得到一个映射值，那么主席树维护这个值就可以了，然后查询就相当于查询区间内不同数的个数。

代码：

#include<bits/stdc++.h>

using namespace std;
const int N =1e5+5;
const int maxnode=2e5+5;
const int sigma_size=30;

int pre[2*N];
int T[N];
int cnode;

struct node
{
    int l,r;
    int ls,rs;
    int sum;
}tr[10000005];

int build(int l,int r)
{
    int cur=++cnode;
    tr[cur].l=l; tr[cur].r=r; tr[cur].sum=0;
    if(l==r) return cur;
    int mid=(l+r)>>1;
    tr[cur].ls=build(l,mid);
    tr[cur].rs=build(mid+1,r);
    return cur;
}

int update(int root,int pos,int val)
{
    int cur=++cnode;
    tr[cur]=tr[root];
    tr[cur].sum+=val;
    if(tr[root].l==pos&&tr[root].r==tr[root].l) return cur;
    int mid=(tr[root].l+tr[root].r)>>1;
    if(pos<=mid) tr[cur].ls=update(tr[root].ls,pos,val);
    else tr[cur].rs=update(tr[root].rs,pos,val);
    return cur;
}

int query(int lrot,int rrot,int l,int r)
{
    if(tr[lrot].l==l&&tr[rrot].r==r){
        return tr[rrot].sum-tr[lrot].sum;
    }
    int mid=(tr[lrot].l+tr[lrot].r)>>1;
    if(r<=mid) return query(tr[lrot].ls,tr[rrot].ls,l,r);
    else if(l>mid) return query(tr[lrot].rs,tr[rrot].rs,l,r);
    else{
        return query(tr[lrot].ls,tr[rrot].ls,l,mid)+query(tr[lrot].rs,tr[rrot].rs,mid+1,r);
    }
}

struct Trie
{
    int ch[maxnode][sigma_size];
    int val[maxnode];
    int num[maxnode];
    int sz; // 结点总数
    void clear()
    {
        sz = 1;
        memset(ch[0], 0, sizeof(ch[0]));
    }
    int idx(char c)
    {
        return c - 'a';
    }

    void insert(const char *s, int id)
    {
        int u = 0, n = strlen(s);
        int RT=T[id-1];
        int cur;
        for(int i = 0; i < n; i++)
        {
            int c = idx(s[i]);
            if(!ch[u][c])
            {
                //cout<<"NNOOOOO"<<endl;
                memset(ch[sz], 0, sizeof(ch[sz]));
                val[sz] = 0;
                num[sz]=0;
                ch[u][c] = sz++;
                u=ch[u][c];
                cur=update(RT,pre[u],1);
                pre[u]=id;
            }
            else{
                //cout<<"YESSSSSS"<<endl;
                u=ch[u][c];
                //cout<<"pre "<<pre[u]<<endl;
                cur=update(RT,pre[u],1);
                pre[u]=id;
            }
            RT=cur;
            num[u]++;
        }
        val[u] = id;
        T[id]=cur;
    }

    bool find(const char *s, int len)
    {
        int u = 0;
        for(int i = 0; i < len; i++)
        {
            if(s[i] == '\0') break;
            int c = idx(s[i]);
            if(!ch[u][c]) break;
            u = ch[u][c];
            if(val[u] != 0) return true; // 找到一个前缀
        }
        return false;
    }

}a;

char str[100005];
int n;

int main()
{
    int z=0;
    while(scanf("%d",&n)!=EOF){
        a.clear();
        z=0;
        cnode=0;
        memset(T,0,sizeof(T));
        memset(pre,0,sizeof(pre));
        T[0]=build(0,2e5+3);

        for(int i=1;i<=n;i++){
            scanf("%s",str);
            a.insert(str,i);
        }
        int q;
        scanf("%d",&q);
        int l,r;
        while(q--)
        {
            scanf("%d %d",&l,&r);
            int tmpl=min((l+z)%n,(r+z)%n)+1;
            int tmpr=max((l+z)%n,(r+z)%n)+1;
            //cout<<"l "<<tmpl<<" r "<<tmpr<<endl;
            if(tmpr<tmpl) swap(tmpl,tmpr);
            int ans=query(T[tmpl-1],T[tmpr],0,tmpl-1);
            printf("%d\n",ans);
            z=ans;
        }
    }
    return 0;
}