[ZOJ4089]Little Sub and Isomorphism Sequences

Time Limit: 3 Seconds
Memory Limit: 65536 KB

Little Sub has a sequence $A_1,A_2,A_3,...,A_N$ . Now he has a problem for you.
Two sequences $X_1,X_2,X_3,...,X_u$ of length $u$ and of $X_1,X_2,X_3,...,X_v$ length $v$ are considered isomorphic when they meet all the following two conditions:
$1. u = v$ ;
$2.$ Define $occ(S,k)$ as the number of times integer $k$ occur in sequence $S$ . For each integer $k$ in $[1,10^9]$ , $occ(X,k)=occ(Y,k)$ always holds.

Now we have $M$ operations for $A$ . and there are two kinds of operations:
$1$ $x$ $y$ : Change $A_x$ to $y$ $(1<=x<=N , 1<=y<=10^9)$ ;
$2$ : Query for the greatest $k(1<=k<N)$ that there exist two integers $i$ and $j$ $(1<=i<j<N-k+1)$ and $A_i,A_{i+1},A_{i+2},...,A_{i+k-1}$ is isomorphic with $A_j,A_{j+1},A_{j+2},...,A_{j+k-1}$ . Specially, if there is no such $k$ , please output “ $-1$ ” (without quotes) instead.

Input

There are multiple test cases. The first line of the input contains an integer $T(1<=T<=5)$ , indicating the number of test cases. For each test case:
The first line ontains two integers $N,M(1<=N,M<=10^5)$ .
The second line contains $N$ integers $A_1,A_2,A_3,...,A_N(1<=A_i<=10^9)$ .
In the following $M$ lines, each line contains one operation. The format is described above.

Output

For each operation 2, output one line containing the answer.

Sample Input

Sample Output

-1
1
2

题意：
给一个序列，有两个操作，第一个操作是单点修改，第二个操作是询问整个序列中最长的两个不重合同构子串的长度。（同构的定义是字串中所有出现的数字，在两个子串中的出现次数都相同）
题解：
我们可以发现，最长不重合同构子串肯定是相互重合的，只有两端是不重合的，那么我们只需要维护 $A[i]$ 中相同数字的最大距离 $w$ 就可以了，答案就是 $w+1$
离散化，开20万个set，存数字 $x$ 的位置信息，然后用线段树维护 $w+1$ ,就可以了。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#define LiangJiaJun main
using namespace std;
set<int>pos[200004];
map<int,int>mert;
int up;
int tr[800004],n,m;
struct als{
    int x,lp;
}a[200004],A[100004];
struct mot{
    int type,x,g,lp;
}work[100004];
inline bool dex(als X,als Y){
    return X.x<Y.x;
}

void update(int x){
     if(pos[x].size()<=1)tr[x+up]=-1;
     else{
        tr[x+up]=(*--pos[x].end())-(*pos[x].begin());
     }
     int now=x+up;now>>=1;
     while(now>=1){
        tr[now]=max(tr[now<<1],tr[now<<1|1]);
        now>>=1;
     }
}
int w33ha(){
    mert.clear();
    scanf("%d%d",&n,&m);
    int cnt=n;
    for(int i=1;i<=n;i++){
        scanf("%d",&A[i].x);
        a[i]=A[i];
    }
    for(int i=1;i<=m;i++){
        scanf("%d",&work[i].type);
        if(work[i].type==1){
            scanf("%d%d",&work[i].g,&work[i].x);
            a[++cnt].x=work[i].x;
        }
    }
    sort(a+1,a+cnt+1,dex);
    a[0].x=0;
    int om=0;
    for(int i=1;i<=cnt;i++){
        if(a[i].x!=a[i-1].x){
            mert[a[i].x]=++om;
            a[i].lp=om;
            pos[om].clear();
        }
    }
    for(int i=1;i<=n;i++)A[i].lp=mert[A[i].x];
    for(int i=1;i<=m;i++){
        if(work[i].type==1)work[i].lp=mert[work[i].x];
    }
    for(int i=1;i<=n;i++){
        pos[A[i].lp].insert(i);
    }
    up=1;
    while(up<cnt+2)up<<=1;
    for(int i=up;i>=1;i--){
        tr[i<<1]=-1;
        tr[i<<1|1]=-1;
    }
    for(int i=1;i<=cnt;i++){
        if(pos[a[i].lp].size()<=1)tr[a[i].lp+up]=-1;
        else{
            tr[a[i].lp+up]=(*--pos[a[i].lp].end())-(*pos[a[i].lp].begin());
        }
    }
    for(int i=up;i>=1;i--)tr[i]=max(tr[i<<1],tr[i<<1|1]);
    for(int i=1;i<=m;i++){
        if(work[i].type==1){
            int g=work[i].g;
            pos[A[g].lp].erase(g);
            pos[work[i].lp].insert(g);
            update(A[g].lp);
            update(work[i].lp);
            A[g].lp=work[i].lp;
            A[g].x=work[i].x;
        }
        else{
            printf("%d\n",tr[1]);
        }
    }
    return 0;
}
int LiangJiaJun(){
    int T;scanf("%d",&T);
    while(T--)w33ha();
    return 0;
}

[ZOJ4089]Little Sub and Isomorphism Sequences

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