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

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

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

$;$
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 $() that there exist two integers and () and is isomorphic with . Specially, if there is no such, please output "-1" (without quotes) instead.$

Input

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 $lines, each line contains one operation. The format is described above.$

Output

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

Sample Input

Sample Output

-1
1
2

题意：给定你个数组，以及一些单点修改，以及询问，每次询问需要求得，最长的字串长度，它在其他位置存在同构。

思路：最长同构子串对总是会重叠的，然后两个子串不重叠的部分必须是同构。那么一定有，最优之一就是“x+公共”与“公共+x”这两个同构最长。

这个不难反证。那么实现就是每种树建立set，然后保存每个set的最远距离即可。

#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 (同构 set)

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