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.
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    const int maxn = 1e3;
    struct segmentTree {
        int l,r;
        int dat;
    } e[4 * maxn];
  • 线段树3个操作
  1. 建树
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    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
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    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
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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
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    #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:
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    #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;
    }

算法

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