SPOJ GSS3 Can you answer these queries III (线段树)

题意:带修 求区间最大连续子段和

题解:我们需要维护的信息有 区间和 区间最大子段和 区间左连续最大子段和 区间右连续最大子段和

　　 然后模拟即可

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e4 + 5;
int n, m;

int a[MAXN];
struct node {
    int sum, ssum, ls, rs;
}E[MAXN << 2];

void pushup(int rt) {
    int lr = rt << 1;
    int rr = rt << 1 | 1;
    E[rt].ssum = E[lr].ssum + E[rr].ssum;
    E[rt].sum = max(E[lr].sum, E[rr].sum);
    E[rt].sum = max(E[rt].sum, E[lr].rs + E[rr].ls);
    E[rt].ls = max(E[lr].ls, E[lr].ssum + E[rr].ls);
    E[rt].rs = max(E[rr].rs, E[rr].ssum + E[lr].rs);
}

void build(int l, int r, int rt) {
    if(l == r) {
        E[rt].sum = a[l];
        E[rt].ls = E[rt].rs = E[rt].ssum = a[l];
        return;
    }

    int mid = l + r >> 1;
    build(l, mid, rt << 1);
    build(mid + 1, r, rt << 1 | 1);
    pushup(rt);
}

void update(int k, int v, int l, int r, int rt) {
    if(l == r) {
        E[rt].ssum = E[rt].sum = E[rt].ls = E[rt].rs = v;
        return;
    }

    int mid = l + r >> 1;
    if(k <= mid) update(k, v, l, mid, rt << 1);
    else update(k, v, mid + 1, r, rt << 1 | 1);
    pushup(rt);
}

node query(int ql, int qr, int l, int r, int rt) {
    if(ql <= l && qr >= r) return E[rt];

    int mid = l + r >> 1;
    if(qr <= mid) return query(ql, qr, l, mid, rt << 1);
    if(ql > mid) return query(ql, qr, mid + 1, r, rt << 1 | 1);
    node ll = query(ql, qr, l, mid, rt << 1);
    node rr = query(ql, qr, mid + 1, r, rt << 1 | 1);
    node res;
    res.ssum = ll.ssum + rr.ssum;
    res.sum = max(ll.sum, rr.sum);
    res.sum = max(res.sum, ll.rs + rr.ls);
    res.ls = max(ll.ls, ll.ssum + rr.ls);
    res.rs = max(rr.rs, rr.ssum + ll.rs);
    return res;
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(1, n, 1);
    scanf("%d", &m);
    for(int i = 1; i <= m; i++) {
        int opt, x, y;
        scanf("%d%d%d", &opt, &x, &y);
        if(opt == 1) printf("%d
", query(x, y, 1, n, 1).sum);
        else update(x, y, 1, n, 1);
    }
    return 0;
}