线段树模板总结

本文只总结一下线段树的易错点（毕竟这玩意我早就会了）

重点：

• 线段树tag的意义
• 线段树开4倍空间的意义

注释里都有。

update函数任务清单

• pushdown
• 递归到子区间
• pushup

query函数任务清单

• pushdown
• 递归到子区间

//线段树不能严格开4倍空间, 要开大于等于MAXN的最小2的幂次的4倍空间
//如：MAXN = 10000, 则要开 16384*4 的空间。
#include<bits/stdc++.h>
using namespace std;

#define ll long long 
const int MAXN = 550005;
ll sum[MAXN], tag[MAXN]; 
//tag表示：这个区间已处理过，但它的子区间还没有。
void pushdown(ll o, ll l, ll r) {
    ll ls, rs, mid;
    if(tag[o] == 0) return;
    ls = o << 1; rs = ls + 1; mid = (l+r) >> 1;
    sum[ls] += tag[o] * (mid - l + 1);
    sum[rs] += tag[o] * (r - mid);
    tag[ls] += tag[o]; tag[rs] += tag[o];
    tag[o] = 0;
}

ll query(ll o, ll l, ll r, ll ql, ll qr) {
    pushdown(o, l, r);
// printf("[%lld,%lld]: %lld
", l, r, sum[o]);
    if(ql <= l && r <= qr) {
        return sum[o];
    }
    ll mid = (l+r)>>1, ret = 0, ls, rs;
    ls = o << 1; rs = ls + 1;
    if(ql <= mid) ret += query(ls, l, mid, ql, qr);
    if(qr > mid)  ret += query(rs, mid+1, r, ql, qr);
printf("%d %d %d ql=%d qr=%d ret= %lld
", l, r, sum[o], ql, qr, ret);

    return ret;
}
void update(ll o, ll l, ll r, ll ql, ll qr, ll v) {
    pushdown(o, l, r);
    ll mid = (l+r)>>1, ls, rs;
    if(ql <= l && r <= qr) {
        sum[o] += (r-l+1) * v;
        tag[o] += v;
        return;
    }
    ls = o << 1; rs = ls + 1;
    if(ql <= mid)  update(ls, l, mid, ql, qr, v);
    if(qr > mid)   update(rs, mid+1, r, ql, qr, v);
    sum[o] = sum[ls] + sum[rs];
}
int a[MAXN], n, q;
signed main(){
    #ifndef ONLINE_JUDGE
    freopen(".in","r",stdin);
    freopen(".out","w", stdout);
    #endif
    scanf("%d %d", &n, &q);
    for(int i = 1; i <= n; i++) {
        ll tmp;
        scanf("%lld", &tmp);
        update(1, 1, n, i, i, tmp);
    }
    for(int i = 1; i <= q; i++) {
        int op, l, r, k;
        scanf("%d%d%d",&op,&l,&r);
        if(op == 1) {
            scanf("%d", &k);
            update(1, 1, n, l, r, k);
        } else {
            printf("%lld
", query(1,1,n,l,r));
        }
    }

    return 0;
}