10.CF515E Drazil and Park 简单线段树

10.CF515E Drazil and Park 简单线段树

给定一个环，$i-i+1$之间的距离给定，每个点有点权$h_i$
每次询问会禁止经过一个区间，求两个点$(s, t)$能够使得$2 * (h_s + h_t) + dis(s, t)$取得最大值
求这个最大值

洛谷传送门：​​CF515E Drazil and Park - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)​​

CF传送门：​​E. Drazil and Park (codeforces.com)​​

题目分析

• 给定一个环，$i-i+1$之间的距离给定，每个点有点权$h_i$
• 每次询问会禁止经过一个区间，求两个点$(s, t)$能够使得$2 * (h_s + h_t) + dis(s, t)$取得最大值
• 求这个最大值

把式子转化一下：

• 对$dis$求前缀和$dd$，则$dis(l, r)=dd[r] - dd[l]$
• 原式化为$dd[r] - dd[l] + 2h[l] + 2h[r] = (dd[r] + 2h[r]) - (dd[l] - 2h[l])$
• 对$dd[r] + 2h[r]$维护区间最大值，对$dd[l] - 2h[l]$维护区间最小值即可

于是就变成了线段树维护区间最大值最小值的过程。在计算过程中对答案取最大值即可。

Code

#include <bits/stdc++.h>
#pragma gcc optimize("O2")
#pragma g++ optimize("O2")
#define int long long
#define endl '\n'
using namespace std;

namespace ffastIO {
    const int bufl = 1 << 15;
    char buf[bufl], *s = buf, *t = buf;
    inline int fetch() {
        if (s == t) { t = (s = buf) + fread(buf, 1, bufl, stdin); if (s == t) return EOF; }
        return *s++;
    }
    inline int read() {
        int a = 0, b = 1, c = fetch();
        while (!isdigit(c))b ^= c == '-', c = fetch();
        while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
        return b ? a : -a;
    }
}

const int N = 2e5 + 10, MOD = 1e9 + 7;
int d[N << 1], h[N << 1], s[N << 1], n, m;

int sum(int l, int r){
    if(l <= r) return h[r] - h[l - 1];
    else return h[l + n] - h[r - 1];
}

int dist(int l, int r){
    if(l <= r) return d[r] - d[l - 1];
    else return d[l + n] - d[r - 1];
}

namespace SegTree{
    #define ls rt << 1
    #define rs rt << 1 | 1
    #define lson ls, l,
    #define rson rs, mid + 1,
    #define tii tuple<int, int, int>
    int maxx[N << 2], minn[N << 2], ans[N << 2];

    inline void push_up(int rt){
        maxx[rt] = max(maxx[ls], maxx[rs]);
        minn[rt] = max(minn[ls], minn[rs]);
        ans[rt] = max({minn[ls] + maxx[rs], ans[ls], ans[rs]});
    }

    void build(int rt, int l, int r){
        if(l == r){
            maxx[rt] = s[l - 1] + h[l];
            minn[rt] = -s[l - 1] + h[l];
            return;
        }
        int mid = l + r >> 1;
        build(lson), build(rson);
        push_up(rt);
    }

    tii query(int rt, int l, int r, int L, int R){
        if(l >= L && r <= R) return tii{maxx[rt], minn[rt], ans[rt]};
        int mid = l + r >> 1; tii ans1, ans2; bool flag1 = false, flag2 = false;
        if(mid >= L) flag1 = true, ans1 = query(lson, L, R);
        if(mid < R) flag2 = true, ans2 = query(rson, L, R);
        if(flag1 && flag2){
            return tii{max(get<0>(ans1), get<0>(ans2)), max(get<1>(ans1), get<1>(ans2)), max({get<1>(ans1) + get<0>(ans2), get<2>(ans1), get<2> (ans2)})};
        }
        else if(flag1) return ans1;
        else return ans2;
    }

}

#define SEGRG 1, 1,

inline void solve(){
    n = ffastIO::read(), m = ffastIO::read();
    for(int i = 1; i <= n; i++) d[i] = d[i + n] = ffastIO::read();
    for(int i = 1; i <= n; i++) h[i] = h[i + n] = ffastIO::read() * 2;
    n *= 2;
    for(int i = 1; i <= n; i++) s[i] = s[i - 1] + d[i];
    // for(int i = 1; i <= n; i++) cout << d[i] << " \n"[i == n * 2];
    // for(int i = 1; i <= n; i++) cout << h[i] << " \n"[i == n * 2];
    SegTree::build(SEGRG);
    for(int i = 1; i <= m; i++){
        int l = ffastIO::read(), r = ffastIO::read();
        if(l <= r) cout << get<2>(SegTree::query(SEGRG, r + 1, l + (n >> 1) - 1)) << endl;
        else cout << get<2>(SegTree::query(SEGRG, r + 1, l - 1)) << endl;
    }
}

signed main(){
    int t = 1; // cin >> t;
    while(t--) solve();
    return 0;
}