数列分块入门 8

 

这题我们只需设个targe[i]表示第i块是否都为一个数,如果是，则targe[i]等于这个数，否则等于-1
然后暴力搞就可以了。以下大佬分析

 

 

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <random>
using namespace std;
#define io ios::sync_with_stdio(0),cin.tie(0)
#define ms(arr) memset(arr,0,sizeof(arr))
#define LD long double
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define inf 1<<30
#define  ull unsigned long long
const int Mod = 10007;
const int maxn = 1e6 + 5;
int read() {
    int x = 0, f = 1;
    char ch = getchar();
    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, m, block;
int opt, l, r, c;
int a[maxn];
int bel[maxn];
int targe[maxn];
void rest(int x)
{
    if (targe[bel[x]] == -1)return;
    for (int i = (bel[x] - 1) * block + 1; i <= min(n, bel[x] * block); i++) {
        a[i] = targe[bel[x]];
    }
    targe[bel[x]] = -1;
}
void search(int L, int R, int val)
{
    int ans = 0;
    rest(L);
    for (int i = L; i <= min(bel[L] * block, R); i++) {
        if (a[i] == val)
            ans++;
        else
            a[i] = val;
    }
    if (bel[L] != bel[R]) {
        rest(R);
        for (int i = (bel[R] - 1) * block + 1; i <= R; i++) {
            if (a[i] == val)
                ans++;
            else
                a[i] = val;
        }
        for (int i = bel[L] + 1; i <= bel[R] - 1; i++) {
            if (targe[i] == -1) {
                for (int j = (i - 1) * block + 1; j <= i * block; j++) {
                    if (a[j] == val)
                        ans++;
                    else
                        a[j] = val;
                }
                targe[i] = val;
            }
            else
            {
                if (targe[i] == val)
                    ans += block;
                else
                    targe[i] = val;
            }
        }
    }
    printf("%d\n", ans);
}
int main()
{
    n = read();
    block = sqrt(n);
    m = n / block;
    if (n % block)m++;
    memset(targe, -1, sizeof(targe));
    for(int i= 1; i <= n; i++) {
        a[i] = read();
        bel[i] = (i - 1) / block + 1;
    }
    for (int i = 1; i <= n; i++) {
        l = read();
        r = read();
        c = read();
        search(l, r, c);
    }
}