• Codeforces 351D Jeff and Removing Periods 线段树


    Jeff and Removing Periods

    题目转换一下就变成询问区间里面有多少个不同数字, 和区间里面有没有等差数列的相同数字。

    离线之后可能分别用树状数组和线段树维护, 线段树维护数字等差数列能延续到的最右端。

    #include<bits/stdc++.h>
    #define LL long long
    #define LD long double
    #define ull unsigned long long
    #define fi first
    #define se second
    #define mk make_pair
    #define PLL pair<LL, LL>
    #define PLI pair<LL, int>
    #define PII pair<int, int>
    #define SZ(x) ((int)x.size())
    #define ALL(x) (x).begin(), (x).end()
    #define fio ios::sync_with_stdio(false); cin.tie(0);
    
    using namespace std;
    
    const int N = 2e5 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 998244353;
    const double eps = 1e-8;
    const double PI = acos(-1);
    
    template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
    template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
    template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
    template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}
    
    struct Bit {
        int a[N];
        void modify(int x, int v) {
            for(int i = x; i < N; i += i & -i)
                a[i] += v;
        }
        int sum(int x) {
            int ans = 0;
            for(int i = x; i; i -= i & -i)
                ans += a[i];
            return ans;
        }
        int query(int L, int R) {
            if(L > R) return 0;
            return sum(R) - sum(L - 1);
        }
    } bit;
    
    #define lson l, mid, rt << 1
    #define rson mid + 1, r, rt << 1 | 1
    
    struct segmentTree {
        int mx[N << 2];
        void modify(int p, int val, int l, int r, int rt) {
            if(l == r) {
                mx[rt] = val;
                return;
            }
            int mid = l + r >> 1;
            if(p <= mid) modify(p, val, lson);
            else modify(p, val, rson);
            mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
        }
        int query(int L, int R, int l, int r, int rt) {
            if(R < l || r < L || R < L) return 0;
            if(L <= l && r <= R) return mx[rt];
            int mid = l + r >> 1;
            return max(query(L, R, lson), query(L, R, rson));
        }
    } Tree;
    
    int n, q, b[N], ans[N];
    int L[N], R[N], Map[N];
    int to[N], dif[N], nex[N];
    
    vector<int> qus[N];
    
    int main() {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
        scanf("%d", &q);
        for(int i = 1; i <= q; i++) {
            scanf("%d%d", &L[i], &R[i]);
            qus[L[i]].push_back(i);
        }
        for(int i = n; i >= 1; i--) {
            if(Map[b[i]]) bit.modify(Map[b[i]], -1);
            bit.modify(i, 1);
            Map[b[i]] = i;
            for(auto& id : qus[i]) ans[id] = bit.query(L[id], R[id]) + 1;
        }
        memset(Map, 0, sizeof(Map));
        for(int i = n; i >= 1; i--) {
            if(Map[b[i]]) {
                int j = Map[b[i]];
                nex[i] = j;
                if(dif[j] == -1) {
                    to[i] = to[j];
                    dif[i] = j - i;
                    Tree.modify(i, to[i], 1, n, 1);
                    Tree.modify(j, 0, 1, n, 1);
                } else {
                    if(j - i == dif[j]) {
                        to[i] = to[j];
                        dif[i] = j - i;
                        Tree.modify(i, to[i], 1, n, 1);
                        Tree.modify(j, 0, 1, n, 1);
                    } else {
                        to[i] = nex[j] - 1;
                        dif[i] = j - i;
                        Tree.modify(i, to[i], 1, n, 1);
                        Tree.modify(j, 0, 1, n, 1);
                    }
                }
            } else {
                nex[i] = n + 1;
                to[i] = n + 1;
                dif[i] = -1;
                Tree.modify(i, to[i], 1, n, 1);
            }
            for(auto& id : qus[i]) {
                if(Tree.query(L[id], R[id], 1, n, 1) >= R[id]) ans[id]--;
            }
            Map[b[i]] = i;
        }
        for(int i = 1; i <= q; i++) printf("%d
    ", ans[i]);
        return 0;
    }
    
    /*
    */
  • 相关阅读:
    TIME_WAIT
    Oracle常用SQL命令
    Linux系统标准化设置:克隆系统后网卡设置、固定IP设置、关闭防火墙和SELINUX、更改主机名
    Linux平台oracle 11g单实例 + ASM存储 安装部署
    Oracle基础维护03-主机、数据库日志收集
    print控制台输出带颜色文字方法
    第二十九节,装饰器
    python 浅析对return的理解
    工欲善其事,必先利其器——持续更新中……
    优秀的学习资源收集:
  • 原文地址:https://www.cnblogs.com/CJLHY/p/10977134.html
Copyright © 2020-2023  润新知