zoukankan      html  css  js  c++  java
  • 2019牛客暑期多校训练营(第四场)- sequence

    单调栈 + 线段树

    先单调栈预处理a中每一个数覆盖的范围,然后用线段树维护b的前缀和,在选i的范围内查询区间值。

    讨论一下a的正负性

    #include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    #define FAST_IO ios::sync_with_stdio(false)
    using namespace std;
    typedef long long LL;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int ret = 0, w = 0; char ch = 0;
        while(!isdigit(ch)){
            w |= ch == '-', ch = getchar();
        }
        while(isdigit(ch)){
            ret = (ret << 3) + (ret << 1) + (ch ^ 48);
            ch = getchar();
        }
        return w ? -ret : ret;
    }
    inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }
    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 = 4000005;
    int n, a[N], b[N], L[N], R[N];
    LL mx[N<<2], mn[N<<2], sum[N];
    
    void calc(){
        stack<int> st;
        a[0] = a[n + 1] = -INF;
        st.push(0);
        for(int i = 1; i <= n; i ++){
            while(!st.empty() && a[st.top()] >= a[i]) st.pop();
            L[i] = st.top();
            st.push(i);
        }
        while(!st.empty()) st.pop();
        st.push(n + 1);
        for(int i = n; i >= 1; i --){
            while(!st.empty() && a[st.top()] >= a[i]) st.pop();
            R[i] = st.top();
            st.push(i);
        }
    }
    
    void push_up(int rt){
        mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
        mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);
    }
    
    void buildTree(int rt, int l, int r){
        if(l == r){
            mx[rt] = mn[rt] = sum[l];
            return;
        }
        int mid = (l + r) >> 1;
        buildTree(rt << 1, l, mid);
        buildTree(rt << 1 | 1, mid + 1, r);
        push_up(rt);
    }
    
    LL queryMax(int rt, int l, int r, int ql, int qr){
        if(l == ql && r == qr){
            return mx[rt];
        }
        int mid = (l + r) >> 1;
        if(qr <= mid) return queryMax(rt << 1, l, mid, ql, qr);
        else if(ql > mid) return queryMax(rt << 1 | 1, mid + 1, r, ql, qr);
        else return max(queryMax(rt << 1, l, mid, ql, mid), queryMax(rt << 1 | 1, mid + 1, r, mid + 1, qr));
    }
    
    LL queryMin(int rt, int l, int r, int ql, int qr){
        if(l == ql && r == qr){
            return mn[rt];
        }
        int mid = (l + r) >> 1;
        if(qr <= mid) return queryMin(rt << 1, l, mid, ql, qr);
        else if(ql > mid) return queryMin(rt << 1 | 1, mid + 1, r, ql, qr);
        else return min(queryMin(rt << 1, l, mid, ql, mid), queryMin(rt << 1 | 1, mid + 1, r, mid + 1, qr));
    }
    
    int main(){
    
        n = read();
        for(int i = 1; i <= n; i ++) a[i] = read();
        for(int i = 1; i <= n; i ++) b[i] = read();
        for(int i = 1; i <= n; i ++) sum[i] = sum[i - 1] + b[i];
        calc();
        buildTree(1, 0, n);
        LL ans = 0;
        for(int i = 1; i <= n; i ++){
            int pl = L[i], pr = i - 1;
            int sl = i, sr = R[i] - 1;
            if(a[i] < 0){
                ans = max(ans, 1LL * a[i] * (queryMin(1, 0, n, sl, sr) - queryMax(1, 0, n, pl, pr)));
            }
            else if(a[i] > 0){
                ans = max(ans, 1LL * a[i] * (queryMax(1, 0, n, sl, sr) - queryMin(1, 0, n, pl, pr)));
            }
        }
        printf("%lld
    ", ans);
        return 0;
    }
    
  • 相关阅读:
    laravel 使用DB 鏈接leftJoin查詢
    checkbox复选框,如何让其勾选时触发一个事件,取消勾选时不触发
    js获取上传图片大小,判断上传图片类型,获取图片真实宽度和高度
    如何查看crontab的日志记录
    linux应用之gcc环境的安装
    laravel 获取上一条insert语句产生的id
    laravel多条件查询(and,or嵌套查询)
    laravel ORM 模型关联 with () 用法
    js实现表单提交 onsubmit
    如何利用jquery来给input添加或删除disabled属性
  • 原文地址:https://www.cnblogs.com/onionQAQ/p/11260417.html
Copyright © 2011-2022 走看看