数列分块入门 8

分块训练

区间染色，统计等于某种颜色的数量。

这里还是用分块，用一个数组记录块内修改的颜色，这样块内查询就可以O(1)染色了，其他区间暴力统计，修改。

注意的是在暴力统计的时候，先要把染色标记push到区间内。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 100005;
ll a[N], col[N];
int n, t, lt[N], rt[N], pos[N];

void calc(int k){
    if(!col[k]) return;
    for(int i = lt[k]; i <= rt[k]; i ++){
        a[i] = col[k];
    }
    col[k] = 0;
}

int solve(int l, int r, ll d){
    int p = pos[l], q = pos[r], ret = 0;
    if(p == q){
        calc(p);
        for(int i = l; i <= r; i ++){
            if(a[i] == d) ret ++;
            a[i] = d;
        }
    }
    else{
        for(int i = p + 1; i <= q - 1; i ++){
            if(col[i]) ret += col[i] == d ? (rt[i] - lt[i] + 1) : 0;
            else{
                for(int j = lt[i]; j <= rt[i]; j ++){
                    if(a[j] == d) ret ++;
                    a[j] = d;
                }
            }
            col[i] = d;
        }
        calc(p);
        for(int i = l; i <= rt[p]; i ++){
            if(a[i] == d) ret ++;
            a[i] = d;
        }
        calc(q);
        for(int i = lt[q]; i <= r; i ++){
            if(a[i] == d) ret ++;
            a[i] = d;
        }
    }
    return ret;
}

int main(){

    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    t = (int)sqrt(n);
    for(int i = 1; i <= t; i ++){
        lt[i] = (i - 1) * t + 1;
        rt[i] = i * t;
    }
    if(rt[t] < n) t ++, lt[t] = rt[t - 1] + 1, rt[t] = n;
    for(int i = 1; i <= t; i ++){
        for(int j = lt[i]; j <= rt[i]; j ++){
            pos[j] = i;
        }
    }
    for(int i = 1; i <= n; i ++){
        int l, r; ll c;
        cin >> l >> r >> c;
        cout << solve(l, r, c) << endl;
    }
    return 0;
}