权值线段树学习笔记

参考博文：

https://www.cnblogs.com/zmyzmy/p/9529234.html

权值线段树：

权值线段树维护数的个数，数组下标代表整个值域，如果太大可以采用离散化。

定义：

struct SegmentTree
{
    int l, r;
    int s; //节点p的s表示这一段值域中数的个数总和
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define s(x) tree[x].s
    #define lson (p<<1)
    #define rson (p<<1|1)
}tree[maxn<<2];

建树：

void build(int p, int l, int r)
{
    l(p) = l, r(p) = r;
    if(l == r)
    {
        //初始化
        return;
    }
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid+1, r);
    //pushup()
}

单点更新：

void change(int p, int x)
{
    if(l(p) == r(p))
    {
        //更新数据
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if(x <= mid) change(lson, x);
    else change(rson, x);
    //pushup()
}

询问整体第\(k\)小：

//询问整个区间第k小
//s(p)代表l(p)到r(p)值域中树的个数总和
int query(int p, int k)
{
    if(l(p) == r(p))
        return l(p); //由于数组下标维护的是值域,直接返回下标
    if(k <= s(lson)) return query(lson, k); //在左子树中
    else return query(rson, k - s(lson)); //在右子树中,感觉和平衡树好像
}

例题1：洛谷 https://www.luogu.org/problem/P1801

链接

思路：

依题意模拟

代码：

#include<bits/stdc++.h>
#include<cstring>
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 10;

int a[maxn], num[maxn], u[maxn];
int n, m, len;

struct SegmentTree
{
    int l, r;
    int s; //节点p的s表示这一段值域中数的个数总和
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define s(x) tree[x].s
    #define lson (p<<1)
    #define rson (p<<1|1)
}tree[maxn<<2];

void build(int p, int l, int r)
{
    l(p) = l, r(p) = r;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid+1, r);
}

void change(int p, int x)
{
    if(l(p) == r(p))
    {
        s(p) += 1;
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if(x <= mid) change(lson, x);
    else change(rson, x);
    s(p) = s(lson) + s(rson);
}

//询问整个区间第k大
//s(p)代表l(p)到r(p)值域中树的个数总和
int query(int p, int k)
{
    if(l(p) == r(p))
        return l(p); //由于数组下标维护的是值域,直接返回下标
    if(k <= s(lson)) return query(lson, k); //在左子树中
    else return query(rson, k - s(lson)); //在右子树中,感觉和平衡树好像
}

int main()
{
    scanf("%d%d", &m, &n);
    for(int i = 1; i <= m; i++)
    {
        scanf("%d", &a[i]);
        num[i] = a[i];
    }
    for(int i = 1; i <= n; i++)
        scanf("%d", &u[i]);
    sort(num + 1, num + 1 + m);
    len = unique(num + 1, num + 1 + m) - num - 1;
    build(1, 1, len);
    int cnt = 0, k = 0;
    while(n != cnt)
    {
        cnt++;
        for(int i = u[cnt-1] + 1; i <= u[cnt]; i++)
        {
            int y = lower_bound(num+1, num+1+len, a[i]) - num;
            //y是a(i)在num里的下标
            change(1, y);
        }
        cout << num[query(1, ++k)] << endl;
    }
    return 0;
}

例题2：洛谷1908：逆序对(权值线段树写法)

链接

题意描述：

求逆序对数目。

思路：

见注释

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 5e5 + 10;
ll ans;
int n, a[maxn], num[maxn], len;

struct SegmentTree
{
    int l, r;
    ll s;
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define lson (p<<1)
    #define rson (p<<1|1)
    #define s(x) tree[x].s
}tree[maxn<<2];

inline void pushup(int p){
    s(p) = s(lson) + s(rson);
}

inline void build(int p, int l, int r)
{
    l(p) = l, r(p) = r;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid + 1, r);
}

inline void change(int p, int x)
{
    if(l(p) == r(p))
    {
        s(p)++;
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if(x <= mid) change(lson, x);
    else change(rson, x);
    pushup(p);
}

ll query(int p, int x)
{
    if(l(p) == r(p)) return s(p);
    int mid = (l(p) + r(p)) >> 1;
    if(x <= mid) return query(lson, x) + s(rson);
    else return query(rson, x);
}


int main()
{
    scanf("%d", &n);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        num[i] = a[i];
    }
    build(1, 1, n);

    sort(num + 1, num + 1 + n);
    len = unique(num + 1, num + 1 + n) - num - 1;

    for(int i = 1; i <= n; i++)
    {
        int p = lower_bound(num + 1, num + 1 + n, a[i]) - num;
        a[i] = p;
    }

    for(int i = 1; i <= n; i++) //枚举每个a(i)作为右端点
    {                           //看树中有多少比他大的数字
        ans += query(1, a[i] + 1); //寻找比当前数大的数字的个数
        //+1是因为要过滤掉等于a(i)的
        change(1, a[i]);  //在权值线段树中加上该节点
    }
    cout << ans << endl;
    return 0;
}

例题3：hdu_4217

链接

题意描述：

给定一个\(1\)到\(n\)的序列。每次操作查询序列第\(k\)小的数字加入答案并拿走这个数字，问最后拿走数字的总和是多少
\(n\leq 3e5\)

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 10;

int T, n, m, k, cas;

struct SegmentTree
{
    int l, r;
    int s;
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define lson (p<<1)
    #define rson (p<<1|1)
    #define s(x) tree[x].s
}tree[maxn<<2];

void pushup(int p){
    s(p) = s(lson) + s(rson);
}

void build(int p, int l, int r)
{
    l(p) = l, r(p) = r;
    if(l == r) {s(p) = 1; return;}
    int mid = (l + r) >> 1;
    build(lson, l, mid);
    build(rson, mid+1, r);
    pushup(p);
}

int query(int p, int k)
{
    if(l(p) == r(p)) return l(p);
    if(k <= s(lson)) return query(lson, k);
    else return query(rson, k - s(lson));
}

void change(int p, int x, int val)
{
    if(l(p) == r(p))
    {
        s(p) = val;
        return;
    }
    int mid = (l(p) + r(p)) >> 1;
    if(x <= mid) change(lson, x, val);
    else change(rson, x, val);
    pushup(p);
}

int main()
{
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &m);
        build(1, 1, n);
        long long ans = 0;
        while(m--)
        {
            scanf("%d", &k);
            int num = query(1, k);
            ans += num;
            change(1, num, 0);
        }
        printf("Case %d: %lld\n", ++cas, ans);
    }
    return 0;
}

来源：https://www.cnblogs.com/zxytxdy/p/11767047.html

标签

线段树

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