The number of colors "National Team"

The meaning of problems

Mo repair team template question

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Thinking

To be repaired on the basis of common Mo Mo team squad on a modified additional maintenance operations, modify or revoke each addition modified according to the time of the query operation.

Achieve quite obvious, with the code it is easy to get to know.

note: See also proved block size here , not repeat them here.

Code

#include <bits/stdc++.h>

using namespace std;

namespace StandardIO {

    template<typename T>inline void read (T &x) {
        x=0;T f=1;char c=getchar();
        for (; c<'0'||c>'9'; c=getchar()) if (c=='-') f=-1;
        for (; c>='0'&&c<='9'; c=getchar()) x=x*10+c-'0';
        x*=f;
    }

    template<typename T>inline void write (T x) {
        if (x<0) putchar('-'),x*=-1;
        if (x>=10) write(x/10);
        putchar(x%10+'0');
    }

}

using namespace StandardIO;

namespace Project {
    
    const int N=200200;
    
    int n,m,block;
    int a[N];
    int qcnt,ucnt;
    struct ask {
        int l,r,t,id;
    } q[N];
    struct upd {
        int x,pre,now;
    } u[N];
    int cnt[1000001],ans[N],res;
    
    inline bool cmp (const ask x,const ask y) {
        return (x.l/block!=y.l/block)?(x.l/block<y.l/block):((x.r/block!=y.r/block)?(x.r/block<y.r/block):x.t<y.t);
    }

    inline void MAIN () {
        read(n),read(m);
        for (register int i=1; i<=n; ++i) 
            read(a[i]);
        for (register int i=1; i<=m; ++i) {
            scanf("\n");
            if (getchar()=='Q') {
                ++qcnt,read(q[qcnt].l),read(q[qcnt].r),q[qcnt].t=ucnt,q[qcnt].id=qcnt;
            } else {
                ++ucnt,read(u[ucnt].x),read(u[ucnt].now),u[ucnt].pre=a[u[ucnt].x],a[u[ucnt].x]=u[ucnt].now;
            }
        }
        for (register int i=ucnt; i; --i) 
            a[u[i].x]=u[i].pre;
        block=ceil(exp((log(n)+log(ucnt))/3)),sort(q+1,q+qcnt+1,cmp);
        int l=1,r=0,t=0;
        for (register int i=1; i<=qcnt; ++i) {
            while (l>q[i].l) res+=!cnt[a[--l]]++;
            while (l<q[i].l) res-=!--cnt[a[l++]];
            while (r<q[i].r) res+=!cnt[a[++r]]++;
            while (r>q[i].r) res-=!--cnt[a[r--]];
            while (q[i].t<t) {
                int x=u[t].x;
                if (l<=x&&x<=r) res-=!--cnt[a[x]];
                a[x]=u[t--].pre;
                if (l<=x&&x<=r) res+=!cnt[a[x]]++;
            }
            while (q[i].t>t) {
                int x=u[++t].x;
                if (l<=x&&x<=r) res-=!--cnt[a[x]];
                a[x]=u[t].now;
                if (l<=x&&x<=r) res+=!cnt[a[x]]++;
            }
            ans[q[i].id]=res;
        }
        for (register int i=1; i<=qcnt; ++i) {
            write(ans[i]),puts("");
        }
    }
    
}

int main () {
//  freopen(".in","r",stdin);
//  freopen(".out","w",stdout);
    Project::MAIN();
}