UV 1400-& amp; quot; Ray, Pass me the dishes! & Amp; quot; (line segment tree)

UV 1400-& quot; Ray, Pass me the dishes! & Quot; (line segment tree)

Link to the question: Ultraviolet 1400-"Ray, Pass me the dishes! "

Given a sequence of n integers, give an answer to m queries. For each query (a, B), find two subscripts x, y makes the continuity from x to y equal to the maximum continuous sum in the range a and B. If multiple solutions exist, x is given priority and y is smaller.

Solution: The line segment tree maintains three line segment values for each node:

• Max_sub: Maximum consecutive intervals and
• Max_prefix: Maximum range continuous prefix and
• Max_suffix: Maximum range continuous suffixes and
  The Build Process maintains three values. You only need to consider max_sub of the left and right subnodes and max_suffix + max_prefix on both sides of the query.

  #include 
      
       #include 
       
        #include using namespace std;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1)+1)typedef long long ll;const int maxn = 500005;const ll INF = 0x3f3f3f3f3f3f3f3f;struct segment {    int l, r;    ll val;    segment (int l = 0, int r = 0, ll val = 0) {        this->set(l, r, val);    }    void set (int l, int r, ll val) {        this->l = l;        this->r = r;        this->val = val;    }    segment operator + (const segment& u) {        return segment(min(l, u.l), max(r, u.r), val + u.val);    }    bool operator < (const segment& u) const {        if (val != u.val)            return val < u.val;        if (l != u.l)            return l > u.l;        return r > u.r;    }};struct Node {    int l, r;    segment max_sub, max_prefix, max_suffix;} node[4*maxn];int N, M;ll arr[maxn], s[maxn];Node seg_push (Node a, Node b) {    Node ret;    ll suml = s[a.r] - s[a.l-1];    ll sumr = s[b.r] - s[b.l-1];    ret.l = a.l;    ret.r = b.r;    ret.max_sub = max(a.max_suffix + b.max_prefix, max(a.max_sub, b.max_sub));    ret.max_prefix = max(a.max_prefix, segment(a.l, a.r, suml) + b.max_prefix);    ret.max_suffix = max(b.max_suffix, a.max_suffix + segment(b.l, b.r, sumr));    return ret;}void build_segTree (int root, int l, int r) {    if (l == r) {        node[root].l = node[root].r = l;        node[root].max_sub.set(l, r, (ll)arr[l]);        node[root].max_prefix.set(l, r, (ll)arr[l]);        node[root].max_suffix.set(l, r, (ll)arr[l]);        return;    }    int mid = (l + r) / 2;    build_segTree(lson(root), l, mid);    build_segTree(rson(root), mid + 1, r);    node[root] = seg_push(node[lson(root)], node[rson(root)]);}Node query (int root, int l, int r) {    if (l <= node[root].l && r >= node[root].r)        return node[root];    int mid = (node[root].l + node[root].r) / 2;    if (l <= mid && r > mid)        return seg_push(query(lson(root), l, r), query(rson(root), l, r));    else if (r <= mid)        return query(lson(root), l, r);    else        return query(rson(root), l, r);}int main () {    int cas = 1;    while (scanf("%d%d", &N, &M) == 2) {        s[0] = 0;        for (int i = 1; i <= N; i++) {            scanf("%lld", &arr[i]);            s[i] = s[i-1] + arr[i];        }        build_segTree(1, 1, N);        printf("Case %d:\n", cas++);        int l, r;        for (int i = 0; i < M; i++) {            scanf("%d%d", &l, &r);            Node u = query(1, l, r);            printf("%d %d\n", u.max_sub.l, u.max_sub.r);        }    }    return 0;}