Codeforces Round #292 (Div. 1) C - Drazil and Park

C - Drazil and Park

每个点有两个值Li 和 Bi，求Li + Rj (i < j) 的最大值，这个可以用线段树巧妙的维护。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;

int n, m;
LL d[N], h[N], sum[N];

struct node {
    LL mx1, mx2, mx3;
    node operator + (const node &rhs) const {
        node ans;
        ans.mx1 = max(mx1, rhs.mx1);
        ans.mx2 = max(mx2, rhs.mx2);
        ans.mx3 = max(mx3, rhs.mx3);
        ans.mx3 = max(ans.mx3, mx1 + rhs.mx2);
        return ans;
    }
} a[N << 2];

void build(int l, int r, int rt) {
    if(l == r) {
        a[rt].mx1 = 2 * h[l] - sum[l - 1];
        a[rt].mx2 = 2 * h[l] + sum[l - 1];
        a[rt].mx3 = -INF;
        return;
    }
    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    a[rt] = a[rt << 1] + a[rt << 1 | 1];
}

node query(int L, int R, int l, int r, int rt) {
    if(l >= L && r <= R) return a[rt];
    int mid = l + r >> 1;
    if(R <= mid) return query(L, R, l, mid, rt << 1);
    if(L > mid) return query(L, R, mid + 1, r, rt << 1 | 1);
    return query(L, R, l, mid, rt << 1) + query(L, R, mid + 1, r, rt << 1 | 1);
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%lld", &d[i]), d[i + n] = d[i];
    for(int i = 1; i <= n; i++) scanf("%lld", &h[i]), h[i + n] = h[i];
    for(int i = 1; i <= 2 * n; i++) sum[i] = sum[i - 1] + d[i];
    build(1, 2 * n, 1);

    while(m--) {
        int L, R; scanf("%d%d", &L, &R);
        L--, R--;
        L = (L - 1 + n) % n;
        R = (R + 1) % n;
        L++, R++;
        swap(L, R);
        if(L > R) R += n;
        printf("%lld\n", query(L, R, 1, 2 * n, 1).mx3);
    }
    return 0;
}

/*
*/