AcWing - 245 - 你能回答这些问题吗 = 线段树

最新推荐文章于 2024-10-26 16:47:14 发布

转载最新推荐文章于 2024-10-26 16:47:14 发布 · 104 阅读

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原文链接：http://www.cnblogs.com/Inko/p/11509727.html

本文介绍了一种使用线段树求解最大连续子序列和的问题，特别关注了至少包含一个元素的约束条件。通过维护从左/右开始的最大连续和，解决了跨越区间的最大和计算问题。代码实现包括构建、更新和查询操作。

https://www.acwing.com/problem/content/246/

一个很有意思的线段树，一般来说要求的最大连续和只需要维护一个从左侧开始/从右侧开始的最大连续和用来跨越区间中点，但是这里居然至少要包含一个元素，所以要进行一些变形。主要是叶子节点里面的各个标记至少要有一个元素。

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

const int MAXN = 500000;
//MAXLSUM和MAXRSUM表示一直从左侧/右侧点开始的最大和，MAXSUM表示无限制的最大和，SUM表示和
ll MAXLSUM[(1 << 20) + 5], MAXRSUM[(1 << 20) + 5], SUM[(1 << 20) + 5], MAXSUM[(1 << 20) + 5];
int A[MAXN + 5];

inline void PushUp(const int &o) {
    MAXLSUM[o] = max(MAXLSUM[o << 1], SUM[o << 1] + max(MAXLSUM[o << 1 | 1], 0ll));
    MAXRSUM[o] = max(MAXRSUM[o << 1 | 1], SUM[o << 1 | 1] +  max(MAXRSUM[o << 1], 0ll));
    MAXSUM[o] = max(max(MAXSUM[o << 1], MAXSUM[o << 1 | 1]), max(max(MAXRSUM[o << 1], 0ll) + MAXLSUM[o << 1 | 1], MAXRSUM[o << 1] + max(MAXLSUM[o << 1 | 1], 0ll)));
    SUM[o] = SUM[o << 1] + SUM[o << 1 | 1];
}

inline void Build(const int &o, const int &l, const int &r) {
    if(l == r) {
        MAXLSUM[o] = MAXRSUM[o] = MAXSUM[o] = SUM[o] = A[l];
        return;
    }
    int mid = l + r >> 1;
    Build(o << 1, l, mid);
    Build(o << 1 | 1, mid + 1, r);
    PushUp(o);
}

inline void Update(const int &o, const int &l, const int &r, const int &x, const int &y) {
    if(l == r) {
        MAXLSUM[o] = MAXRSUM[o] = MAXSUM[o] = SUM[o] = A[l] = y;
        return;
    }
    int mid = l + r >> 1;
    if(x <= mid)
        Update(o << 1, l, mid, x, y);
    else
        Update(o << 1 | 1, mid + 1, r, x, y);
    PushUp(o);
}

const ll INF = 1e18;

inline ll QueryMAXLSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXLSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(ql <= mid)
        res = QueryMAXLSUM(o << 1, l, mid, ql, qr);
    if(qr >= mid + 1)
        res = max(res, SUM[o << 1] + max(QueryMAXLSUM(o << 1 | 1, mid + 1, r, ql, qr), 0ll));
    return res;
}

inline ll QueryMAXRSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXRSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(qr >= mid + 1)
        res = QueryMAXRSUM(o << 1 | 1, mid + 1, r, ql, qr);
    if(ql <= mid)
        res = max(res, SUM[o << 1 | 1] + max(QueryMAXRSUM(o << 1, l, mid, ql, qr), 0ll));
    return res;
}

inline ll QueryMAXSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(ql <= mid)
        res = QueryMAXSUM(o << 1, l, mid, ql, qr);
    if(qr >= mid + 1)
        res = max(res, QueryMAXSUM(o << 1 | 1, mid + 1, r, ql, qr));
    if(ql <= mid && qr >= mid + 1) {
        ll tmpL = QueryMAXRSUM(o << 1, l, mid, ql, qr);
        ll tmpR = QueryMAXLSUM(o << 1 | 1, mid + 1, r, ql, qr);
        res = max(res, max(tmpL + max(tmpR, 0ll), max(tmpL, 0ll) + tmpR));
    }
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &A[i]);
    }
    Build(1, 1, n);
    while(m--) {
        int op, x, y;
        scanf("%d%d%d", &op, &x, &y);
        if(op == 1) {
            if(x > y)
                swap(x, y);
            printf("%lld\n", QueryMAXSUM(1, 1, n, x, y));
        } else {
            Update(1, 1, n, x, y);
        }
    }
}

实际上已经保证至少有一个元素之后一部分max可以去掉。

inline void PushUp(const int &o) {
    MAXLSUM[o] = max(MAXLSUM[o << 1], SUM[o << 1] + max(MAXLSUM[o << 1 | 1], 0ll));
    MAXRSUM[o] = max(MAXRSUM[o << 1 | 1], SUM[o << 1 | 1] +  max(MAXRSUM[o << 1], 0ll));
    MAXSUM[o] = max(max(MAXSUM[o << 1], MAXSUM[o << 1 | 1]), max(max(MAXRSUM[o << 1], 0ll) + MAXLSUM[o << 1 | 1], MAXRSUM[o << 1] + max(MAXLSUM[o << 1 | 1], 0ll)));
    SUM[o] = SUM[o << 1] + SUM[o << 1 | 1];
}

因为MAXLSUM[o]>=SUM[o]，且MAXRSUM[o]>=SUM[o]，所以可以去掉前两个max操作使得常数变小。
而第三个max操作，假如有某个前缀/后缀和是负数的话，那么最大值肯定是落在其中一个区间的。

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

const int MAXN = 500000;
//MAXLSUM和MAXRSUM表示一直从左侧/右侧点开始的最大和，MAXSUM表示无限制的最大和，SUM表示和
ll MAXLSUM[(1 << 20) + 5], MAXRSUM[(1 << 20) + 5], SUM[(1 << 20) + 5], MAXSUM[(1 << 20) + 5];
int A[MAXN + 5];

inline void PushUp(const int &o) {
    MAXLSUM[o] = max(MAXLSUM[o << 1], SUM[o << 1] + MAXLSUM[o << 1 | 1]);
    MAXRSUM[o] = max(MAXRSUM[o << 1 | 1], SUM[o << 1 | 1] +  MAXRSUM[o << 1]);
    MAXSUM[o] = max(max(MAXSUM[o << 1], MAXSUM[o << 1 | 1]), max(MAXRSUM[o << 1] + MAXLSUM[o << 1 | 1], MAXRSUM[o << 1] + MAXLSUM[o << 1 | 1]));
    SUM[o] = SUM[o << 1] + SUM[o << 1 | 1];
}

inline void Build(const int &o, const int &l, const int &r) {
    if(l == r) {
        MAXLSUM[o] = MAXRSUM[o] = MAXSUM[o] = SUM[o] = A[l];
        return;
    }
    int mid = l + r >> 1;
    Build(o << 1, l, mid);
    Build(o << 1 | 1, mid + 1, r);
    PushUp(o);
}

inline void Update(const int &o, const int &l, const int &r, const int &x, const int &y) {
    if(l == r) {
        MAXLSUM[o] = MAXRSUM[o] = MAXSUM[o] = SUM[o] = A[l] = y;
        return;
    }
    int mid = l + r >> 1;
    if(x <= mid)
        Update(o << 1, l, mid, x, y);
    else
        Update(o << 1 | 1, mid + 1, r, x, y);
    PushUp(o);
}

const ll INF = 1e18;

inline ll QueryMAXLSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXLSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(ql <= mid)
        res = QueryMAXLSUM(o << 1, l, mid, ql, qr);
    if(qr >= mid + 1)
        res = max(res, SUM[o << 1] + QueryMAXLSUM(o << 1 | 1, mid + 1, r, ql, qr));
    return res;
}

inline ll QueryMAXRSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXRSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(qr >= mid + 1)
        res = QueryMAXRSUM(o << 1 | 1, mid + 1, r, ql, qr);
    if(ql <= mid)
        res = max(res, SUM[o << 1 | 1] + QueryMAXRSUM(o << 1, l, mid, ql, qr));
    return res;
}

inline ll QueryMAXSUM(const int &o, const int &l, const int &r, const int &ql, const int &qr) {
    if(ql <= l && r <= qr) {
        return MAXSUM[o];
    }
    int mid = l + r >> 1;
    ll res = -INF;
    if(ql <= mid)
        res = QueryMAXSUM(o << 1, l, mid, ql, qr);
    if(qr >= mid + 1)
        res = max(res, QueryMAXSUM(o << 1 | 1, mid + 1, r, ql, qr));
    if(ql <= mid && qr >= mid + 1) {
        ll tmpL = QueryMAXRSUM(o << 1, l, mid, ql, qr);
        ll tmpR = QueryMAXLSUM(o << 1 | 1, mid + 1, r, ql, qr);
        res = max(res, tmpL + tmpR);
    }
    return res;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, m;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &A[i]);
    }
    Build(1, 1, n);
    while(m--) {
        int op, x, y;
        scanf("%d%d%d", &op, &x, &y);
        if(op == 1) {
            if(x > y)
                swap(x, y);
            printf("%lld\n", QueryMAXSUM(1, 1, n, x, y));
        } else {
            Update(1, 1, n, x, y);
        }
    }
}

转载于:https://www.cnblogs.com/Inko/p/11509727.html