树状数组模板(改点求段 / 该段求点 / 改段求段)

1. 改点求段(单点更新, 区间求和)

代码:

#include <iostream>
using namespace std;

const int MAXN = 1e5 + 10;
int tree[MAXN], n;

int lowbit(int x){//返回 pow(2, k),其中k为末尾0的个数, 即返回最低位1的值
    return x & -x;
}

void add(int x, int d){//将d累加到tree数组对应位置
    while(x <= n){
        tree[x] += d;
        x += lowbit(x);
    }
}

int sum(int x){//向上求和, 返回[1, x]所有元素的和
    int ans = 0;
    while(x > 0){
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}

int main(void){
    int x;
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> x;
        add(i, x);
    }
    for(int i = 1; i <= n; i++){
        cout << sum(i) << endl;
    }
    return 0;
}

2. 改点求段(区间更新, 单点求值)

 用一个数组 d 存储目标数组 a 中相邻元素的差值, 即 i > 1 时, d[i] = a[i] - a[i - 1] ; i == 1 时, d[i] = a[i] .

那么有 a[i] = d[1] + ... + d[i] .若要将 a 数组区间 [l, r] 的元素都加上 key, 显然只需令 d[l] += key, d[r + 1] -= key 即可.

显然只要用树状数组维护一下 d 数组即可.

代码:

#include <iostream>
#include <stdio.h>
using namespace std;

const int MAXN = 1e5 + 10;
int  a[MAXN], tree[MAXN], d[MAXN], n;

int lowbit(int x){
    return x & -x;
}

void add(int x, int ad){
    while(x <= n){
        tree[x] += ad;
        x += lowbit(x);
    }
}

int sum(int x){//sum(x)为a[x]的值
    int ans = 0;
    while(x > 0){
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}

int main(void){
    int x, q, l, r;
    cin >> n;
    for(int i = 1; i <= n; i++){
        scanf("%d", &a[i]);
        if(i == 1) d[i] = a[i];
        else d[i] = a[i] - a[i - 1];
    }
    for(int i = 1; i <= n; i++){
        add(i, d[i]);
    }
    cin >> q;
    while(q--){
        cin >> l >> r >> x;//将区间[l, r]的元素都加上x
        add(l, x);
        add(r + 1, -x);
        for(int i = 1; i <= n; i++){
            cout << sum(i) << " ";
        }
        cout << endl;
    }
}

3. 改段求段

 与 2 中类似, 先开一个差分数组 d

那么有:

a1 + a2 + ... + an

= d1 + (d1 + d2) + ... + (d1 + d2 + ... + dn)

= n * d1 + (n - 1) * d2 + ... + dn

= n * (d1 + d2 + ... + dn) - (0 * d1 + 1 * d2 + ... (n - 1) * dn)

再令 c[i] = ( i - 1) * di

那么原式可化简为:

 n * (d1 + d2 + ... + dn) - (c1 + c2 + ... cn)

显然对于 d 和 c 数组求和可以用树状数组解决. 而由 2 可知 a 数组区间修改则只需要对 d 和 c 数组对应做单点修改即可.

例题: http://codevs.cn/problem/1082/

题意: 中文题诶~

思路: 树状数组区间更新区间求和模板

代码:

#include <iostream>
#include <stdio.h>
#define ll long long
using namespace std;

const int MAXN = 2e5 + 10;
ll a[MAXN], d[MAXN], c[MAXN];//d[i]=a[i]-a[i-1], c[i]=(i-1)*d[i]
int n;

int lowbit(int x){
    return x & -x;
}

void add(ll *tree, int x, ll ad){
    while(x <= n){
        tree[x] += ad;
        x += lowbit(x);
    }
}

ll sum(ll *tree, int x){
    ll ans = 0;
    while(x > 0){
        ans += tree[x];
        x -= lowbit(x);
    }
    return ans;
}

int main(void){
    ll ad;
    int q, op, l, r;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++){
        scanf("%lld", &a[i]);
        add(d, i, a[i] - a[i - 1]);
        add(c, i, (i - 1) * (a[i] - a[i - 1]));
    }
    scanf("%d", &q);
    while(q--){
        scanf("%d", &op);
        if(op == 1){
            scanf("%d%d%lld", &l, &r, &ad);
            add(d, l, ad);
            add(d, r + 1, -ad);
            add(c, l, (l - 1) * ad);
            add(c, r + 1, -ad * r);
        }else{
            scanf("%d%d", &l, &r);
            ll sum1 = (l - 1) * sum(d, l - 1) - sum(c, l - 1);
            ll sum2 = r * sum(d, r) - sum(c, r);
            printf("%lld
", sum2 - sum1);
        }
    }
    return 0;
}