SP1716 GSS3 - Can you answer these queries III

SP1716 GSS3 - Can you answer these queries

也真是服了，浪费几分钟来搞这种题目。

直接线段树维护一下端点信息即可，具体来说就是左右端点的权值最大值，答案还有区间权值和。

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#include <bits/stdc++.h>
using namespace std;

//#define Fread
//#define Getmod

#ifdef Fread
char buf[1 << 21], *iS, *iT;
#define gc() (iS == iT ? (iT = (iS = buf) + fread (buf, 1, 1 << 21, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
#define getchar gc
#endif // Fread

template <typename T>
void r1(T &x) {
	x = 0;
	char c(getchar());
	int f(1);
	for(; c < '0' || c > '9'; c = getchar()) if(c == '-') f = -1;
	for(; '0' <= c && c <= '9';c = getchar()) x = (x * 10) + (c ^ 48);
	x *= f;
}

template <typename T,typename... Args> inline void r1(T& t, Args&... args) {
    r1(t);  r1(args...);
}

//#define int long long
const int maxn = 2e5 + 5;
const int maxm = maxn << 1;

int n, m;
int a[maxn];

struct Node {
    int mxl, mxr, val, sum;
}t[maxn << 2];

Node merge(Node u, Node v) {
    Node res;
    res.mxl = max(u.mxl, u.sum + v.mxl);
    res.mxr = max(v.mxr, v.sum + u.mxr);
    res.sum = u.sum + v.sum;
    res.val = max(max(u.val, v.val), u.mxr + v.mxl);
    return res;
}

struct Seg {
    #define ls (p << 1)
    #define rs (p << 1 | 1)
    #define mid ((l + r) >> 1)
    void build(int p,int l,int r) {
        if(l == r) return t[p].mxl = t[p].mxr = t[p].val = t[p].sum = a[l], void();
        build(ls, l, mid), build(rs, mid + 1, r);
        t[p] = merge(t[ls], t[rs]);
    }

    void change(int p,int l,int r,int pos) {
        if(l == r) return t[p].mxl = t[p].mxr = t[p].val = t[p].sum = a[l], void();
        if(pos <= mid) change(ls, l, mid, pos);
        else change(rs, mid + 1, r, pos);
        t[p] = merge(t[ls], t[rs]);
    }

    Node Ask(int p,int l,int r,int ll,int rr) {
        if(ll <= l && r <= rr) return t[p];
        if(ll <= mid && rr <= mid) return Ask(ls, l, mid, ll, rr);
        else if(mid < rr && mid < ll) return Ask(rs, mid + 1, r, ll, rr);
        else return merge(Ask(ls, l, mid, ll, rr), Ask(rs, mid + 1, r, ll, rr));
    }

}T;

signed main() {
//    freopen("S.in", "r", stdin);
//    freopen("S.out", "w", stdout);
    int i, j;
    r1(n);
    for(i = 1; i <= n; ++ i) r1(a[i]);
    T.build(1, 1, n);
    r1(m);
    for(i = 1; i <= m; ++ i) {
        int opt, l, r;
        r1(opt, l, r);
        if(opt == 0) {
            a[l] = r;
            T.change(1, 1, n, l);
        }
        else {
            printf("%d\n", T.Ask(1, 1, n, l, r).val);
        }
    }
	return 0;
}