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线段树模版1

作者: 作者的头像   没有AC千万不能睡觉 ,  2024-05-10 10:22:26 ,  所有人可见 ,  阅读 5


1


不引用懒惰标记的方法,会超时

#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 1e6 + 10;
int nums[N];
typedef long long ll;
struct Tree {
    int l, r;
    ll sum;
    int tag;
}tree[N<<2];
void pushup(int u) {
    tree[u].sum = tree[u << 1].sum + tree[u << 1 | 1].sum;
}
void build(int u, int l, int r) {
    if (l == r) {
        tree[u] = { l,r,nums[l] };
        return;
    }
    tree[u] = { l,r };
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
}
void pushdown(int u, int ln, int rn) {
    if (tree[u].tag) {
        tree[u << 1].tag += tree[u].tag;
        tree[u << 1 | 1].tag += tree[u].tag;

        tree[u << 1].sum += tree[u].tag * ln;
        tree[u << 1 | 1].sum += tree[u].tag * rn;
        tree[u].tag = 0;
    }
}
ll query(int u, int L, int R) {
    int l = tree[u].l, r = tree[u].r;
    int mid = (l + r) >> 1;
    if (l >= L && r <= R) {
        return tree[u].sum;
    }
    pushdown(u, mid - l + 1, r - mid);//有懒惰标记时要下推标记 
    ll sum = 0;
    if (L <= mid) sum += query(u << 1, L, R);
    if (R > mid) sum += query(u << 1 | 1, L, R);
    return sum;
}

void modify_span(int u, int L,int R, int val) {
    int l = tree[u].l, r = tree[u].r;
    int mid = (l + r) >> 1;
    if (l == r) {
        tree[u].sum += val;
        return;
    }

    if(L <= mid)modify_span(u << 1, L, R, val);
    if(R > mid) modify_span(u << 1 | 1, L, R, val);

    pushup(u);
}
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> nums[i];
    }
    build(1,1,n);
    while (m--) {//操作数
        int op;
        cin>>op;
        if (op == 1) {
            int x, y, k;
            cin >> x >> y >> k;
            modify_span(1, x, y, k);
        }
        else {
            int x, y;
            cin >> x >> y;
            cout << query(1, x, y) << endl;
        }
    }
    return 0;
}


记得sum、query要开long long 否则会WA

下面将介绍一种引入一种不超时的方法,时间复杂度变为O(nlogn)

#define _CRT_SECURE_NO_WARNINGS 1
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
const int N = 1e6 + 10;
int nums[N];
typedef long long ll;
struct Tree {
    int l, r;
    ll sum;
    int tag;
}tree[N<<2];
void pushup(int u) {
    tree[u].sum = tree[u << 1].sum + tree[u << 1 | 1].sum;
}
void build(int u, int l, int r) {
    if (l == r) {
        tree[u] = { l,r,nums[l] };
        return;
    }
    tree[u] = { l,r };
    int mid = (l + r) >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
}
void pushdown(int u, int ln, int rn) {
    if (tree[u].tag) {
        tree[u << 1].tag += tree[u].tag;
        tree[u << 1 | 1].tag += tree[u].tag;

        tree[u << 1].sum += tree[u].tag * ln;
        tree[u << 1 | 1].sum += tree[u].tag * rn;
        tree[u].tag = 0;
    }
}
ll query(int u, int L, int R) {
    int l = tree[u].l, r = tree[u].r;
    int mid = (l + r) >> 1;
    if (l >= L && r <= R) {
        return tree[u].sum;
    }
    pushdown(u, mid - l + 1, r - mid);//有懒惰标记时要下推标记 
    ll sum = 0;
    if (L <= mid) sum += query(u << 1, L, R);
    if (R > mid) sum += query(u << 1 | 1, L, R);
    return sum;
}

void modify_span(int u, int L,int R, int val) {
    int l = tree[u].l, r = tree[u].r;
    int mid = (l + r) >> 1;
    if (l >= L && r <= R) {
        tree[u].sum += (r - l + 1) * val;
        tree[u].tag += val;
        pushdown(u, mid - l + 1, r - mid);
        return;
    }

    pushdown(u, mid - l + 1, r - mid);
    if(L <= mid)modify_span(u << 1, L, R, val);
    if(R > mid) modify_span(u << 1 | 1, L, R, val);

    pushup(u);
}
int main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        cin >> nums[i];
    }
    build(1,1,n);
    while (m--) {//操作数
        int op;
        cin>>op;
        if (op == 1) {
            int x, y, k;
            cin >> x >> y >> k;
            modify_span(1, x, y, k);
        }
        else {
            int x, y;
            cin >> x >> y;
            cout << query(1, x, y) << endl;
        }
    }
    return 0;
}

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