Little Sub has a sequence . Now he has a problem for you.

Two sequences of length and of length are considered isomorphic when they meet all the following two conditions:

1. ;
2. Define as the number of times integer occur in sequence . For each integer in , always holds.

Now we have operations for . and there are two kinds of operations:

• 1 x y: Change to (, );
• 2: Query for the greatest () that there exist two integers and () and is isomorphic with . Specially, if there is no such , please output "-1" (without quotes) instead.

There are multiple test cases. The first line of the input contains an integer (), indicating the number of test cases. For each test case:

The first line ontains two integers .

The second line contains integers ().

In the following lines, each line contains one operation. The format is described above.

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

1
3 5
1 2 3
2
1 3 2
2
1 1 2
2

-1
1
2

題意：給定你個數組，以及一些單點修改，以及詢問，每次詢問需要求得，最長的字串長度，它在其他位置存在同構。

思路：最長同構子串對總是會重疊的，然後兩個子串不重疊的部分必須是同構。 那麽一定有，最優之一就是“x+公共”與“公共+x”這兩個同構最長。

這個不難反證。 那麽實現就是每種樹建立set，然後保存每個set的最遠距離即可。

#include<bits/stdc++.h>
#define rep(i,w,v) for(int i=w;i<=v;i++)
using namespace std;
const int maxn=200010;
int a[maxn],b[maxn],c[maxn],x[maxn],y[maxn],tot;
set<int>s[maxn];
multiset<int>S;
multiset<int>::iterator it;
int main()
{
    int T,N,M,ans;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&N,&M); tot=0;
        rep(i,1,N) scanf("%d",&a[i]),b[++tot]=a[i];
        rep(i,1,M){
            scanf("%d",&c[i]);
            if(c[i]==1) {
                scanf("%d%d",&x[i],&y[i]);
                b[++tot]=y[i];
            }
        }
        sort(b+1,b+tot+1);
        tot=unique(b+1,b+tot+1)-(b+1);
        S.clear(); rep(i,1,tot) s[i].clear();
        rep(i,1,N) {
            a[i]=lower_bound(b+1,b+tot+1,a[i])-b;
            s[a[i]].insert(i);
        }
        rep(i,1,tot) if(!s[i].empty())
          S.insert(*(--s[i].end())-*s[i].begin());
        rep(i,1,M){
            if(c[i]==2){
                ans=-1;
                if(!S.empty()) ans=*(--S.end());
                if(ans==0) ans=-1;
                printf("%d\n",ans);
            }
            else {
                it=S.find(*(--s[a[x[i]]].end())-*s[a[x[i]]].begin());
                s[a[x[i]]].erase(x[i]);
                S.erase(it);
                if(!s[a[x[i]]].empty()) S.insert(*(--s[a[x[i]]].end())-*s[a[x[i]]].begin());

                y[i]=lower_bound(b+1,b+tot+1,y[i])-b; a[x[i]]=y[i];
                if(!s[y[i]].empty())  S.erase(S.find(*(--s[y[i]].end())-*s[y[i]].begin()));
                s[y[i]].insert(x[i]);
                S.insert(*(--s[y[i]].end())-*s[y[i]].begin());
            }
        }
    }
    return 0;
}

ZOJ - 4089 :Little Sub and Isomorphism Sequences (同構 set)