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  • 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
    ", query(x, y));
            } else {
                update(x, y);
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/yohaha/p/5095483.html
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