线段树（区间求和、插入、查询最大最小值）-- 模板

题目：
给你一个数组A[1::n]，初始时每个元素都为零。对数组完成一些操作：
第一种可能，给你两个数p 和x（1<= p <= n），把数组的第p 个元素替换为x，即A[p] =x.
第二种可能，给你两个数L 和R（1 <=L <= R <= n），请给出A[L]; A[L + 1]; : : : ; A[R] 这几个数中去掉一个最大值和一个最小值后剩下的数的和是多少.

题解：
单点修改、区间查询的线段树可以在O(log n) 的时间内维护区间的和、最大值、最小值.
此题用三个线段树维护各个区间的和、最大值、最小值，对每个查询输出对应区间的sum . max . min 即可.

#include
#include
using namespace std;
typedef long long ll;

const int maxn = 1e6 + 5;
struct node {
    int left, right;
    int  mx, mn;
    ll sum;
}rt[maxn * 4];

int tl(int x) { return x * 2; }
int tr(int x) { return x * 2 + 1; }

void pushup(int x) {
    rt[x].sum = rt[tl(x)].sum + rt[tr(x)].sum;
    rt[x].mx = max(rt[tl(x)].mx, rt[tr(x)].mx);
    rt[x].mn = min(rt[tl(x)].mn, rt[tr(x)].mn);
}

void build(int x,int L,int R) {
    rt[x].left = L; rt[x].right = R;
    if (L == R) {
        rt[x].sum = rt[x].mn = rt[x].mx = 0;
        return;
    }
    int mid = L + R >> 1;
    build(tl(x), L, mid); build(tr(x), mid + 1, R);
    pushup(x);
}

void upd(int x, int pos, ll val) {
    if (rt[x].left == rt[x].right) {
        rt[x].sum = rt[x].mn = rt[x].mx = val;
        return;
    }
    int mid = rt[x].left + rt[x].right >> 1;
    if (pos <= mid)upd(tl(x), pos, val);
    else upd(tr(x), pos, val);
    pushup(x);
}

ll qury(int x, int L, int R) {
    if (L == rt[x].left&&R == rt[x].right) {
        return rt[x].sum;
    }
    int mid = rt[x].left + rt[x].right >> 1;
    if (R <= mid) return qury(tl(x), L, R);
    else if (L > mid) return qury(tr(x), L, R);
    else return qury(tl(x), L, mid) + qury(tr(x), mid + 1, R);
}

ll fmin(int x, int L, int R) {
    if (L == rt[x].left&&R == rt[x].right) {
        return rt[x].mn;
    }
    int mid = rt[x].left + rt[x].right >> 1;
    ll ret = 0x3fffffffffffff;
    if (R <= mid) return min(ret, fmin(tl(x), L, R));
    else if (L > mid) return min(ret, fmin(tr(x), L, R));
    else return min(fmin(tl(x), L, mid), fmin(tr(x), mid + 1, R));
}

ll fmax(int x, int L, int R) {
    if (L == rt[x].left&&R == rt[x].right) {
        return rt[x].mx;
    }
    int mid = rt[x].left + rt[x].right >> 1;
    ll ret = -0x3ffffffffffffff;
    if (R <= mid) return max(ret, fmax(tl(x), L, R));
    else if (L > mid) return max(ret, fmax(tr(x), L, R));
    else return max(fmax(tl(x), L, mid), fmax(tr(x), mid + 1, R));
}

int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    build(1, 1, n);
    while (m--) {
        int o, x, y;
        scanf("%d%d%d", &o, &x, &y);
        if (o == 0) upd(1, x, y);
        else printf("%lld\n", qury(1, x, y) - fmax(1, x, y) - fmin(1, x, y));
    }
    return 0;
}

转载于:https://www.cnblogs.com/romaLzhih/p/9489838.html