UVA - 1400"Ray, Pass me the dishes!"(线段树)

UVA - 1400"Ray, Pass me the dishes!"(线段树)

题目链接

题目大意：给你N个数字，要求你动态的给出L到R之间，X>= L && Y<=R,使得X，Y这段的连续和是LR之间的最大连续和，如果有多解，输出X小的，接着是Y小的。

解题思路:结点保存三个附加线段，max_sub, max_suffix, max_prefix.对于每次查询最大的连续和要不出现在左子树的max_sub, 要不就是右子树的max_sub, 要不就是左子树的max_suffix + 右子树的max_prefix.然后每次查询的时候都返回一个新的节点，是新的控制范围和在这个范围内的对应的三个线段值。求max_prefix和max_suffix的时候要注意。

代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int N = 5e5 + 5;#define lson(x) ((x)<<1)#define rson(x) (((x)<<1) + 1)ll A[N], S[N];struct Segment {    ll v;    int l, r;    Segment(int l = 0, int r = 0, ll v = 0) {        this->l = l;        this->r = r;        this->v = v;    }    Segment operator + (const Segment& a) const{        Segment ans;        ans.l = min (l, a.l);        ans.r = max (r, a.r);        ans.v = v + a.v;        return ans;    }    bool operator < (const Segment &a) const {        if (v == a.v) {            if (l == a.l)                return r > a.r;            return l > a.l;        }        return v < a.v;    }};struct Node {    int l, r;    Segment max_sub, max_prefix, max_suffix;     void set (int l, int r, Segment max_sub, Segment max_prefix, Segment max_suffix) {            this->l = l;            this->r = r;            this->max_sub = max_sub;            this->max_prefix = max_prefix;            this->max_suffix = max_suffix;    }}node[4 * N];Node Seg_merge(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 (int u, int l, int r) {    if (l == r) {             Segment max_sub(l, r, A[l]);        Segment max_prefix(l, r, A[l]);        Segment max_suffix(l, r, A[l]);        node[u].set(l, r, max_sub, max_prefix, max_suffix);    } else {        int m = (l + r) / 2;                    build(lson(u), l, m);        build(rson(u), m + 1, r);        node[u] = Seg_merge(node[lson(u)], node[rson(u)]);    }        }Node Query (int u, int ql, int qr) {    if (ql <= node[u].l && qr >= node[u].r)        return node[u];    int m = (node[u].l + node[u].r) / 2;    if (ql > m)        return Query (rson(u), ql, qr);    else if (qr <= m)        return Query (lson(u), ql, qr);    else        return Seg_merge(Query(lson(u), ql, qr), Query(rson(u), ql, qr));}int n, m;int main () {    int cas = 0;    int l, r;    Node ans;    while (scanf ("%d%d", &n, &m) != EOF) {        S[0] = 0;        for (int i = 1; i <= n; i++) {            scanf ("%lld", &A[i]);            S[i] = S[i - 1] + A[i];        }        printf ("Case %d:\n", ++cas);        build(1, 1, n);        for (int i = 0; i < m; i++) {            scanf ("%d%d", &l, &r);            ans = Query(1, l, r);            printf ("%d %d\n", ans.max_sub.l, ans.max_sub.r);        }    }    return 0;}