经典颜色问题推荐博文

https://www.cnblogs.com/tyner/p/11519506.html

https://www.cnblogs.com/tyner/p/11616770.html

https://www.cnblogs.com/tyner/p/11620894.html

题意

https://www.luogu.org/problem/P4113

求一段区间中超过出现两次及以上的元素种类

分析

和其他的没啥区别，维护nxt[x], 和nxt[ nxt[x] ]即可， 还是考虑移动左端点对区间答案的影响

#include<cstdio>
#include<algorithm>
using namespace std;
#define lowbit(x) (x&-x)
const int MAX = 2000000+99; 
inline int read() {
    char ch = getchar(); int f = 1, x = 0;
    while(ch<'0' || ch>'9') {if(ch=='-') f = -1; ch = getchar();}
    while(ch>='0' && ch<='9') {x = x*10+ch-'0'; ch = getchar();}
    return x*f;
}

int n,c,m;
int nxt[MAX], lst[MAX], nnxt[MAX];
int arr[MAX], t[MAX];

struct node{
    int l, r, id, ans;
}cmd[MAX];
bool cmp1(node a, node bb) { return a.l < bb.l;}
bool cmp2(node a, node bb) { return a.id < bb.id;}

void add(int x, int k) {while(x <= n) t[x] += k, x += lowbit(x);}
int query(int x) {
    int res = 0;
    while(x) {res += t[x], x -= lowbit(x);}
    return res;
}

void pre() {
    n = read(), c = read(), m = read();
    for(int i = 1; i <= n; i++) arr[i] = read();
    for(int i = n; i >= 1; i--) {
        nxt[i] = lst[arr[i]];
        lst[arr[i]] = i;
    }
    
    for(int i = 1; i <= n; i++) nnxt[i] = nxt[nxt[i]]; 
    for(int i = 1; i <= c; i++) if(lst[i] && nxt[lst[i]]) add(nxt[lst[i]], 1);//颜色数为C 
    //分清n,m 
    for(int i = 1; i <= m; i++) cmd[i].l=read(), cmd[i].r=read(), cmd[i].id = i;
    sort(cmd+1, cmd+1+m, cmp1);
}

void solve() {
    int pos = 1;
    for(int l = 1; l <= n; l++) {
        while(cmd[pos].l == l) {
            cmd[pos].ans = query(cmd[pos].r)-query(cmd[pos].l-1);
            ++pos;
        }
        if(nxt[l]) add(nxt[l], -1);
        if(nnxt[l]) add(nnxt[l], 1);
    }
    
    sort(cmd+1, cmd+1+m, cmp2);
    for(int i = 1; i <= m; i++) printf("%d\n", cmd[i].ans);
}

int main() {
    pre();
    solve();
    return 0;
}