线段树+lazy标记

• 线段树基于分治思想的二叉树,用于区间信息统计
  特点:

1. 每个节点代表区间
2. 唯一根节点,也就是全部区间
3. 叶节点是长度为1的子区间,也就是所代表数组上的一个点

• 存线段树的结构体数组长度一般是 线段树所代表数组长度的4倍.

const int maxn = 1e3;
struct segmentTree {
    int l,r;
    int dat;
} e[4 * maxn];

• 线段树3个操作

1. 建树

void build(int p,int l,int r) {
    e[p].l = l; e[p].r = r;
    if (l == r) {
        e[p].dat = a[l]; // 存原数组的值
        return ;
    }
    int mid = (l+r)/2;
    build(p*2,l,mid); //递归建左子树
    build(p*2+1,mid+1,r); //右子树
    e[p].dat = max(e[p*2].dat,e[p*2+1].dat); //从下到上(回溯)更新区间节点的最大值
}

2. 修改某节点值

void change(int p,int index,int val) {
    if (e[p].l == e[p].r) {
        e[p].dat = val;
        return ;
    }
    int mid = (e[p].l + e[p].r)/2;
    if (index <= mid) change(p*2,index,val);
    else change(p*2+1,index,val);
    e[p].dat = max(e[p*2].dat,e[p*2+1].dat);
}

3. 询问区间最值

int ask(int p,int l,int r) {
    if (l<=e[p].l && e[p].r <= r) return e[p].dat;
    int mid = (e[p].l + e[p].r)/2;
    int val = -(1<<30);
    if (l <= mid) val = max(val,ask(p*2,l,r));
    if (r > mid) val = max(val,ask(p*2+1,l,r));
    return val;
}

线段树例题

HDU 1166

• 此线段树存的是区间和,不是区间最值

#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 2e5+5;

int a[maxn];
int n;

struct segmentTree {
    int l,r;
    int dat;
} e[4 * maxn];

void build(int p,int l,int r) {
    e[p].l = l; e[p].r = r;
    if (l == r) {
        e[p].dat = a[l]; // 存原数组的值
        return ;
    }
    int mid = (l+r)/2;
    build(p*2 , l , mid); //递归建左子树
    build(p*2+1,mid+1,r); //右子树
    e[p].dat = e[p*2].dat + e[p*2+1].dat; //从下到上(回溯)更新区间节点的最大值
}
void change(int p,int index,int val) {
    if (e[p].l == e[p].r) {
        e[p].dat += val;
        return ;
    }
    int mid = (e[p].l + e[p].r)/2;
    if (index <= mid) change(p*2,index,val);
    else change(p*2+1,index,val);
    e[p].dat = e[p*2].dat + e[p*2+1].dat; 
}

int ask(int p,int l,int r) {
    if (l <= e[p].l && e[p].r <= r) return e[p].dat;
    int mid = (e[p].l + e[p].r)/2;
    int val = 0;
    if (l <= mid) val += ask(p*2  ,l,r); // l,r 代表询问的区间 
    if (r > mid)  val += ask(p*2+1,l,r);
    return val;
}

int main() {
//	freopen("a.txt","r",stdin);
	int cnt=0;
	int times; scanf("%d",&times);
	while (times--) {
		scanf("%d",&n);
		memset(e,0,sizeof e);
		memset(a,0,sizeof a);
		for (int i=1; i<=n; i++) scanf("%d",&a[i]);
		build(1,1,n);
		char ch[10];
		printf("Case %d:\n",++cnt);
		while (scanf("%s",ch) && ch[0]!='E') {
			int l,r; scanf("%d%d",&l,&r);
			if (ch[0] == 'Q') 
				printf("%d\n",ask(1,l,r));
			 else if (ch[0] == 'A') 
				change(1,l,r);
			 else if (ch[0] == 'S') 
				change(1,l,-r);
		}
	}
	return 0;
}

---

• 2019/12/15 17:18:15
• 上面线段树写的这么挫,难怪线段树像白学了一样.一直手写不出来,这次重新写一下
• 区间修改+区间查询:

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int maxn = 4e5+10;
int a[maxn/4],left_[maxn],right_[maxn];
ll add[maxn],sum[maxn];

void pushup(int p) { sum[p] = sum[p<<1] + sum[p<<1|1]; }

void pushdown(int p) {
    if (add[p]) {
        sum[p<<1]   += add[p]*(right_[p<<1] - left_[p<<1] + 1);
        sum[p<<1|1] += add[p]*(right_[p<<1|1] - left_[p<<1|1] + 1);
        add[p<<1]   += add[p];
        add[p<<1|1] += add[p];
        add[p] = 0;
    }
}

void build(int p,int l,int r) {
    left_[p]=l,right_[p]=r;
    if (l == r) {
        sum[p] = a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(p<<1,l,mid);
    build(p<<1|1,mid+1,r);
    pushup(p);
}

void change(int p,int l,int r,int val) {
    if (l<=left_[p] && right_[p]<=r) {
        sum[p] += (ll)val * (right_[p] - left_[p] + 1);
        add[p] += val;
        return ;
    }
    // 要向下继续找,则下传标记
    pushdown(p);
    int mid = (left_[p] + right_[p]) >> 1;
    if (l <= mid)
        change(p<<1,l,r,val);
    if (mid < r)
        change(p<<1|1,l,r,val);
    pushup(p);
}

ll query(int p,int l,int r) {
    if (l<=left_[p] && right_[p]<=r)
        return sum[p];
    pushdown(p);
    ll ans = 0;
    int mid = (left_[p] + right_[p]) >> 1;
    if (l <= mid)
        ans += query(p<<1,l,r);
    if (mid < r)
        ans += query(p<<1|1,l,r);
    return ans;
}

int n,m;

int main() {
    [](){
        ::ios::sync_with_stdio(false);
        ::cin.tie(0);
        ::cout.tie(0);
    }();
    cin >> n >> m;
    for (int i=1; i<=n; ++i)
        cin >> a[i];
    build(1,1,n);
    while(m--) {
        int index,x,y,k;
        cin >> index;
        if (index == 1) {
            cin >> x >> y >> k;
            change(1,x,y,k);
        } else {
            cin >> x >> y;
            cout << query(1,x,y) << "\n";
        }
    }
    return 0;
}