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;
}

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