线段树题目

- 题单
- - 线段树1
  - TJOI2009开关
  - 逆序对
  - 忠诚
  - 无聊的数列
  - 扶苏的问题
  - 线段树2
  - [COCI2010-2011#6]STEP
  - 三元上升子序列
  - 色板游戏
  - 小白逛公园
  - 方差
  - [yLOI2019]棠梨煎雪
  - 上帝造题的七分钟2/花神游历各国
  - [SCOI2010]序列操作
  - Points

题单

线段树1

区间相加，区间求和

#include <bits/stdc++.h>
using namespace std;

const int N = 100010;

typedef long long LL;

int n, m;
int w[N];
struct Node {
    int l, r;
    LL sum, add;
}tr[N << 2];

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

void pushdown(int u) {
    auto& root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.add) {
        left.add += root.add, left.sum += (LL)(left.r - left.l + 1) * root.add;
        right.add += root.add, right.sum += (LL)(right.r - right.l + 1) * root.add;
        root.add = 0;
    }
}

void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, w[l], 0};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        // 注意这里
        pushup(u);
    }
}

void modify(int u, int l, int r, int d) {
    if (l <= tr[u].l && r >= tr[u].r) {
        // [l, r]完全包含u对应的区间
        // 处理sum和懒标记
        tr[u].sum += (LL)d * (tr[u].r - tr[u].l + 1);
        tr[u].add += d;
    } else {
        // 注意这里
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r, d);
        if (r > mid) modify(u << 1 | 1, l, r, d);
        // 注意这里
        pushup(u);
    }
}

LL query(int u, int l, int r) {
    if (l <= tr[u].l && r >= tr[u].r) return tr[u].sum;
    
    // 注意这里
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    LL sum = 0;
    if (l <= mid) sum += query(u << 1, l, r);
    if (r > mid) sum += query(u << 1 | 1, l, r);
    return sum;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", w + i);
    
    build(1, 1, n);
    
    int op, l, r, d;
    
    while (m--) {
        scanf("%d", &op);
        if (op == 1) {
            scanf("%d%d%d", &l, &r, &d);
            modify(1, l, r, d);
        } else {
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(1, l, r));
        }
    }
    return 0;
}

TJOI2009开关

一排灯初始是关闭的，每次可翻转一个区间的等的状态，每次询问一个区间内亮的等的个数。相当于是区间改状态，区间求和。
sum表示当前节点的开着的灯的个数，tag表示操作了多少次.

#include <bits/stdc++.h>
using namespace std;
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif
 
const int N = 100010;
 
typedef long long LL;
 
int n, m;
int w[N];
struct Node {
    int l, r;
	// sum当前区间开着的灯的个数 tag标记:初始为0，操作一次变为1
    int sum, tag;
}tr[N << 2];
 
void pushup(int u) {
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}
 
void pushdown(int u) {
    auto& root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.tag) {
        left.tag ^=1, left.sum = (left.r - left.l + 1)- left.sum;
        right.tag ^= 1, right.sum = (right.r - right.l + 1) - right.sum;
        root.tag = 0;
    }
}
 
void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, 0, 0};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        // 注意这里
        pushup(u);
    }
}
 
void modify(int u, int l, int r) {
    if (l <= tr[u].l && r >= tr[u].r) {
        // [l, r]完全包含u对应的区间
        // 处理sum和懒标记
        tr[u].sum = (tr[u].r - tr[u].l + 1) - tr[u].sum;
        tr[u].tag ^= 1;
    } else {
        // 注意这里
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r);
        if (r > mid) modify(u << 1 | 1, l, r);
        // 注意这里
        pushup(u);
    }
}
 
LL query(int u, int l, int r) {
    if (l <= tr[u].l && r >= tr[u].r) {
		// debug(tr[u].l, tr[u].r, tr[u].sum);
		return tr[u].sum;
	}
    
    // 注意这里
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    LL sum = 0;
    if (l <= mid) sum += query(u << 1, l, r);
    if (r > mid) sum += query(u << 1 | 1, l, r);
    return sum;
}
 
int main() {
    scanf("%d%d", &n, &m);
    build(1, 1, n);
    
    int op, l, r, d;
    
    while (m--) {
        scanf("%d", &op);
        if (op == 0) {
            scanf("%d%d", &l, &r);
            modify(1, l, r);
        } else {
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(1, l, r));
        }
    }
    return 0;
}

逆序对

忠诚

区间求最小值.

#include <bits/stdc++.h>
using namespace std;
 
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif
 
typedef long long LL;

const int N = 200010;

struct Node
{
    int l, r;
    int v;  // 区间[l, r]中的最小值
}tr[N * 4];

int w[N];
void pushup(int u)  // 由子节点的信息，来计算父节点的信息
{
    tr[u].v = min(tr[u << 1].v, tr[u << 1 | 1].v);
}

void build(int u, int l, int r)
{
	
	if (l == r) tr[u] = {l, r, w[l]};
	else {
		tr[u] = {l, r};
		int mid = l + r >> 1;
		build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
		pushup(u);
	}
}

int query(int u, int l, int r)
{
    if (tr[u].l >= l && tr[u].r <= r) return tr[u].v; // 树中节点，已经被完全包含在[l, r]中了

    int mid = tr[u].l + tr[u].r >> 1;
    int v = INT_MAX;
    if (l <= mid) v = query(u << 1, l, r);
    if (r > mid) v = min(v, query(u << 1 | 1, l, r));

    return v;
}

// 本题不需要这个函数，这个函数是单点修改
void modify(int u, int x, int v)
{
    if (tr[u].l == x && tr[u].r == x) tr[u].v = v;
    else
    {
        int mid = tr[u].l + tr[u].r >> 1;
        if (x <= mid) modify(u << 1, x, v);
        else modify(u << 1 | 1, x, v);
        pushup(u);
    }
}


int main()
{
    int m, n;
    scanf("%d%d", &m, &n);
	for (int i = 1; i <= m; i++) scanf("%d", w + i);
    build(1, 1, m);

    int x, y;
    while (n--)
    {
        scanf("%d%d", &x, &y);
		printf("%d ", query(1, x, y));
    }

    return 0;
}

无聊的数列

区间加等差数列。
转化为维护原序列的差分数列，那么区间 $[l, r]$ 加等差数列等价于在差分数列的[l]加首项, $[l + 1, r]$ 加公差， $[r]$ 处减去末项. 转化为区间修改，区间求和的问题。

#include <bits/stdc++.h>
using namespace std;
 
const int N = 100010;
 
typedef long long LL;
 
int n, m;
int w[N];
struct Node {
    int l, r;
    LL sum, add;
}tr[N << 2];
 
void pushup(int u) {
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}
 
void pushdown(int u) {
    auto& root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
    if (root.add) {
        left.add += root.add, left.sum += (LL)(left.r - left.l + 1) * root.add;
        right.add += root.add, right.sum += (LL)(right.r - right.l + 1) * root.add;
        root.add = 0;
    }
}
 
void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, w[l], 0};
    else {
        tr[u] = {l, r};
        int mid = l + r >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        // 注意这里
        pushup(u);
    }
}
 
void modify(int u, int l, int r, int d) {
    if (l <= tr[u].l && r >= tr[u].r) {
        // [l, r]完全包含u对应的区间
        // 处理sum和懒标记
        tr[u].sum += (LL)d * (tr[u].r - tr[u].l + 1);
        tr[u].add += d;
    } else {
        // 注意这里
        pushdown(u);
        int mid = tr[u].l + tr[u].r >> 1;
        if (l <= mid) modify(u << 1, l, r, d);
        if (r > mid) modify(u << 1 | 1, l, r, d);
        // 注意这里
        pushup(u);
    }
}
 
LL query(int u, int l, int r) {
    if (l <= tr[u].l && r >= tr[u].r) return tr[u].sum;
    
    // 注意这里
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    LL sum = 0;
    if (l <= mid) sum += query(u << 1, l, r);
    if (r > mid) sum += query(u << 1 | 1, l, r);
    return sum;
}
 
int main() {
    scanf("%d%d", &n, &m);
	//原数组
    for (int i = 1; i <= n; i++) scanf("%d", w + i);
	// 差分数组
	for (int i = n; i > 1; i--) w[i] = w[i] - w[i - 1];
    
    build(1, 1, n);
    
    int op, l, r, k, d;
    
    while (m--) {
        scanf("%d", &op);
        if (op == 1) {
            scanf("%d%d%d%d", &l, &r, &k, &d);
			// 差分数组 w[l] += k
            modify(1, l, l, k);
			// 差分数组 [l+1, r] += d
            if (l + 1 <= r) modify(1, l + 1, r, d);
			// 差分数组 w[r+1]-=末项 也就是 k+d*(r-l)
            if (r < n) modify(1, r + 1, r + 1, -(k + 1LL * d * (r - l)));
        } else {
            scanf("%d", &r);
            printf("%lld\n", query(1, 1, r));
        }
    }
    return 0;
}