bzoj 2120 : 数颜色 分块

题目链接

给一个序列， 两种操作， 一种是将x位置的数改为y， 一种是询问[l, r]之间有多少种不同的数， 数的范围<1e6。

 

分块， 对于每个数， 记录它前面的和他相同的数的位置， 如果pre[i]<l， 那么ans++， 具体看代码......

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int n, m, block, q, a[10005], b[1000005], last[1000005], pre[1000005], pos[10005];
void reset(int x) {
    int l = (x-1)*block+1, r = min(x*block, n);
    for(int i = l; i<=r; i++) {
        pre[i] = b[i];
    }
    sort(pre+l, pre+r+1);
}
void build() {
    for(int i = 1; i<=n; i++) {
        b[i] = last[a[i]];
        last[a[i]] = i;
        pos[i] = (i-1)/block+1;
    }
    for(int i = 1; i<=m; i++)
        reset(i);
}
void update(int p, int val) {
    for(int i = 1; i<=n; i++) {
        last[a[i]] = 0;
    }
    a[p] = val;
    for(int i = 1; i<=n; i++) {
        int t = b[i];
        b[i] = last[a[i]];
        if(t != b[i])
            reset(pos[i]);
        last[a[i]] = i;
    }
}
int find(int x, int pos) {
    int l = (x-1)*block+1, r = min(n, x*block);
    int first = l;
    while(l<=r) {
        int mid = l+r>>1;
        if(pre[mid]<pos)
            l = mid+1;
        else
            r = mid-1;
    }
    return l-first;
}
int query(int l, int r) {
    int ans = 0;
    if(pos[l] == pos[r]) {
        for(int i = l; i<=r; i++) {
            if(b[i]<l)
                ans++;
        }
    } else {
        for(int i = l; i<=(block*pos[l]); i++) {
            if(b[i]<l)
                ans++;
        }
        for(int i = block*(pos[r]-1)+1; i<=r; i++) {
            if(b[i]<l)
                ans++;
        }
    }
    for(int i = pos[l]+1; i<pos[r]; i++)
        ans += find(i, l);
    return ans;
}
int main()
{
    scanf("%d%d", &n, &q);
    for(int i = 1; i<=n; i++)
        scanf("%d", &a[i]);
    block = (int)(sqrt(n)+log(2*n)/log(2));
    if(n%block)
        m = n/block+1;
    else
        m = n/block;
    build();
    char ch;
    int x, y;
    while(q--) {
        scanf(" %c%d%d", &ch, &x, &y);
        if(ch == 'Q') {
            printf("%d\n", query(x, y));
        } else {
            update(x, y);
        }
    }
    return 0;
}