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\n", ans);
    return 0;
}

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转载自www.cnblogs.com/onionQAQ/p/11260417.html