BZOJ 2743: [HEOI2012]采花 [树状数组 | 主席树]

题意:

查询区间中出现次数$>2$的颜色个数

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一眼主席树，区间中$l \le last[i] \le r$的个数减去$l \le last[last[i]] \le r$的个数，搞两颗主席树来做

然后就T了

因为bzoj上数据是1e6....

还是离线树状数组吧....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lc(x) t[x].l
#define rc(x) t[x].r
const int N=1e6+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, k, Q, a[N], last[N], pos[N], l, r;
struct ChairTree{
    struct meow{int l, r, sum;} t[N*20];
    int sz, root[N];
    void ins(int &x, int l, int r, int p) {
        t[++sz] = t[x]; x=sz;
        t[x].sum++;
        if(l==r) return;
        int mid = (l+r)>>1;
        if(p<=mid) ins(t[x].l, l, mid, p);
        else ins(t[x].r, mid+1, r, p);
    }
    int que(int x, int y, int l, int r, int ql, int qr) {
        if(ql<=l && r<=qr) return t[y].sum - t[x].sum;
        else {
            int mid=(l+r)>>1, ans=0;
            if(ql<=mid) ans += que(lc(x), lc(y), l, mid, ql, qr);
            if(mid<qr)  ans += que(rc(x), rc(y), mid+1, r, ql, qr);
            return ans;
        }
    }
}C1, C2;
int main() {
    freopen("in","r",stdin);
    n=read(); k=read(); Q=read();
    for(int i=1; i<=n; i++) a[i]=read(), last[i] = pos[a[i]], pos[a[i]] = i;
    for(int i=1; i<=n; i++) {
        C1.root[i] = C1.root[i-1], C1.ins(C1.root[i], 0, n, last[i]);
        C2.root[i] = C2.root[i-1], C2.ins(C2.root[i], 0, n, last[last[i]]);
    }
    for(int i=1; i<=Q; i++) 
        l=read(), r=read(), 
        printf("%d\n", C1.que(C1.root[l-1], C1.root[r], 0, n, l, r) - C2.que(C2.root[l-1], C2.root[r], 0, n, l, r) );
}

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define lc(x) t[x].l
#define rc(x) t[x].r
const int N=1e6+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, k, Q, a[N], last[N], pos[N], l, r;
int c[N];
inline void add(int p, int v) {if(p!=0) for(;p<=n;p+=(p&-p)) c[p]+=v;}
inline int sum(int p) {int ans=0; for(;p;p-=(p&-p)) ans+=c[p]; return ans;}
struct meow{
    int l, r, qid;
    bool operator <(const meow &a) const {return r<a.r;}
}q[N];
int ans[N], now;
int main() {
    freopen("in","r",stdin);
    n=read(); k=read(); Q=read();
    for(int i=1; i<=n; i++) a[i]=read();
    for(int i=1; i<=Q; i++) l=read(), r=read(), q[i]=(meow){l, r, i};
    sort(q+1, q+1+Q);
    now = 1;
    for(int i=1; i<=n; i++) { 
        last[i] = pos[a[i]]; pos[a[i]] = i;
        add(last[i], 1); add(last[last[i]], -1); 
        while(q[now].r == i) ans[q[now].qid] = sum(i) - sum(q[now].l-1), now++;
    }
    for(int i=1; i<=Q; i++) printf("%d\n", ans[i]);
}