线段树

  • 简介
    线段树是一棵二叉树，树中的每一个结点表示了一个区间[a,b]。每一个叶子节点表示了一个单位区间。对于每一个非叶结点所表示的结点[a,b]，其左儿子表示的区间为[a,(a+b)/2]，右儿子表示的区间为[(a+b)/2,b]。

    线段树模型

  • 线段树需要注意的几点
    1、线段树空间一般要开到数据规模的4~5倍（不卡数据尽可能多开）
    2、lson、rson几个变量尽可能位运算吧（省时），想偷懒就用#define 好了
    3、build函数（下文会介绍）中，if (l == r) sum[n] = a[l] 或sum[n] = a[r]注意不是sum[n] = a[n]
    4、下文几个常用函数中，标记‘*’的可根据需要修改

  • 几个线段树的常用函数
//build（建树用）函数的伪代码 /
build (int n, int l, int r){
    size[n] = r - l + 1;
    if (l == r){
        sum[n] = a[l]; return ;
    }
    int mid = (l + r) >> 1;
    build (n << 1, l, mid);
    build (n << 1 | 1, mid + 1, r);
    sum[n] = sum[n << 1] + sum[n << 1 | 1];//*
}

//updata （区间修改）伪代码/
down (int n){
    sum[n << 1] += lazy[n] * size[n << 1];
    sum[n << 1 | 1] += lazy[n] * size[n << 1 | 1];//*
    /上述两步是区间求和的代码
    求最大值应为
    sum[n << 1] += lazy[n];
    sum[n << 1 | 1] += lazy[n];/
    lazy[n << 1] += lazy[n], lazy[n << 1 | 1] += lazy[n], lazy[n] = 0;
}

updata (int n, int l, int r, int xl, int xr, int v){
    if (l != r) down(n);/即在解决单点修改问题 --> 单点不需要down/
    if (l == xl && r == xr){
        sum[n] += v * size[n], lazy[n] += v;//*
        return ;
    }
    int mid = (l + r) >> 1;
    if (xr <= mid) updata(n << 1, l, mid, xl, xr, v);
    else if (xl > mid) updata(n << 1 | 1, mid + 1, r, xl, xr, v);
    else updata(n << 1, l, mid, xl, mid, v), updata(n << 1 | 1, mid + 1, r, mid + 1, xr, v);
    sum[n] = sum[n << 1] + sum[n << 1 | 1];//*
}

//query （区间查询“sum”函数）伪代码 /
int query (int n, int l, int r, int xl, int xr){
    if (l != r) down(n);
    if (xl <= l && r <= xr) return sum[n];
    int mid = (l + r) >> 1;
    if (xr <= mid) return query(n << 1, l, mid, xl, xr);
    else if (xl > mid) return query(n << 1 | 1, mid + 1, r, xl, xr);
    else return query(n << 1, l, mid, xl, mid) + query(n << 1 | 1, mid + 1, r, mid + 1, xr);
}
#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ls (n << 1)
#define rs (n << 1 | 1)
using namespace std;

int num, m, bj;
int a[1000100];
int size[1000100];
int lazy[1000100];
int sum[1000100];
int n, v, xl, xr;

void build(int n, int l, int r){
    size[n] = r - l + 1;
    if (l == r){
        sum[n] = a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
    sum[n] = sum[ls] + sum[rs];
}

void down(int n){
    sum[ls] += lazy[n] * size[ls];
    sum[rs] += lazy[n] * size[rs];
    lazy[ls] += lazy[n], lazy[rs] += lazy[n], lazy[n] = 0;
}

void add(int n, int l, int r, int xl, int xr, int v){
    if (l != r) down(n);
    if (xl == l && r == xr){
        sum[n] += v * size[n];
        lazy[n] += v;
        return ;
    }
    int mid = (l + r) >> 1;
    if (xr <= mid) add(ls, l, mid, xl, xr, v);
    else if (xl > mid) add(rs, mid + 1, r, xl, xr, v);
    else add(ls, l, mid, xl, mid, v), add(rs, mid + 1, r, mid + 1, xr, v);
    sum[n] = sum[ls] + sum[rs];
}

int query(int n, int l, int r, int xl, int xr){
    if (l != r) down(n);
    if (xl <= l && r <= xr) return sum[n];
    int mid = (l + r) >> 1;
    if (xr <= mid) return query(ls, l, mid, xl, xr);
    else if (xl > mid) return query(rs, mid + 1, r, xl, xr);
    else return query(ls, l, mid, xl, mid) + query(rs, mid + 1, r, mid + 1, xr);
}

int main()
{
    scanf("%d", &num);
    for (int i = 1; i <= num; ++i)    scanf("%d", &a[i]);
    build(1, 1, num);
    scanf("%d", &m);
    for (int i = 1; i <= m; ++i){
        scanf("%d", &bj);
        if (bj == 1) scanf("%d %d", &n, &v), add(1, 1, num, n, n, v);
        else scanf("%d %d", &xl, &xr), printf("%d\n", query(1, 1, num, xl, xr));
    }
    return 0;
}

Codevs 1081线段树练习2

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ls (n << 1)
#define rs (n << 1 | 1)
#define L 500000
using namespace std;

int num, m, bj;
int a[L];
int size[L];
int lazy[L];
int sum[L];
int n, v, l, r, j;

int gi(){
    char cj = getchar();
    int ans = 0;
    while (cj > '9' || cj < '0') cj = getchar();
    while (cj >= '0' && cj <= '9') ans = ans * 10 + cj - '0', cj = getchar();
    return ans;
}

void build(int n, int l, int r){
    size[n] = r - l + 1;
    if (l == r){
        sum[n] = a[l];
        return ;
    }
    int mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
    sum[n] = sum[ls] + sum[rs];
}

void down(int n){
    sum[ls] += lazy[n] * size[ls];
    sum[rs] += lazy[n] * size[rs];
    lazy[ls] += lazy[n], lazy[rs] += lazy[n], lazy[n] = 0;
}

void add(int n, int l, int r, int xl, int xr, int v){
    if (l != r) down(n);
    if (xl == l && r == xr){
        sum[n] += v * size[n];
        lazy[n] += v;
        return ;
    }
    int mid = (l + r) >> 1;
    if (xr <= mid) add(ls, l, mid, xl, xr, v);
    else if (xl > mid) add(rs, mid + 1, r, xl, xr, v);
    else add(ls, l, mid, xl, mid, v), add(rs, mid + 1, r, mid + 1, xr, v);
    sum[n] = sum[ls] + sum[rs];
}

int query(int n, int l, int r, int xl, int xr){
    if (l != r) down(n);
    if (xl <= l && r <= xr) return sum[n];
    int mid = (l + r) >> 1;
    if (xr <= mid) return query(ls, l, mid, xl, xr);
    else if (xl > mid) return query(rs, mid + 1, r, xl, xr);
    else return query(ls, l, mid, xl, mid) + query(rs, mid + 1, r, mid + 1, xr);
}

int main()
{
    num = gi();
    for (int i = 1; i <= num; ++i)    a[i] = gi();
    build(1, 1, num);
    m = gi();
    for (int i = 1; i <= m; ++i){
        scanf("%d", &bj);
        if (bj == 1) scanf("%d %d %d", &l, &r, &v), add(1, 1, num, l, r, v);
        else scanf("%d", &j), printf("%d\n", query(1, 1, num, j, j));
    }
    return 0;
}

Codevs 1082线段树练习3

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ls (n << 1)
#define rs (n << 1 | 1)
#define L 1000000
using namespace std;

long long num, m, bj;
long long a[L];
long long size[L];
long long lazy[L];
long long sum[L];
long long n, v, l, r, j, k;

long long gi(){
    char cj = getchar();
    long long ans = 0;
    while (cj > '9' || cj < '0') cj = getchar();
    while (cj >= '0' && cj <= '9') ans = ans * 10 + cj - '0', cj = getchar();
    return ans;
}

void build(long long n, long long l, long long r){
    size[n] = r - l + 1;
    if (l == r){
        sum[n] = a[l];
        return ;
    }
    long long mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
    sum[n] = sum[ls] + sum[rs];
}

void down(long long n){
    sum[ls] += lazy[n] * size[ls];
    sum[rs] += lazy[n] * size[rs];
    lazy[ls] += lazy[n], lazy[rs] += lazy[n], lazy[n] = 0;
}

void add(long long n, long long l, long long r, long long xl, long long xr, long long v){
    if (l != r) down(n);
    if (xl == l && r == xr){
        sum[n] += v * size[n];
        lazy[n] += v;
        return ;
    }
    long long mid = (l + r) >> 1;
    if (xr <= mid) add(ls, l, mid, xl, xr, v);
    else if (xl > mid) add(rs, mid + 1, r, xl, xr, v);
    else add(ls, l, mid, xl, mid, v), add(rs, mid + 1, r, mid + 1, xr, v);
    sum[n] = sum[ls] + sum[rs];
}

long long query(long long n, long long l, long long r, long long xl, long long xr){
    if (l != r) down(n);
    if (xl <= l && r <= xr) return sum[n];
    long long mid = (l + r) >> 1;
    if (xr <= mid) return query(ls, l, mid, xl, xr);
    else if (xl > mid) return query(rs, mid + 1, r, xl, xr);
    else return query(ls, l, mid, xl, mid) + query(rs, mid + 1, r, mid + 1, xr);
}

int main()
{
    num = gi();
    for (int i = 1; i <= num; ++i)    a[i] = gi();
    build(1, 1, num);
    m = gi();
    for (int i = 1; i <= m; ++i){
        scanf("%lld", &bj);
        if (bj == 1) scanf("%lld %lld %lld", &l, &r, &v), add(1, 1, num, l, r, v);
        else scanf("%lld %lld", &j, &k), printf("%lld\n", query(1, 1, num, j, k));
    }
    return 0;
}

CJOJ P2162 - 线段树-magictree

线段树求和板子题

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define ls (n << 1)
#define rs (n << 1 | 1)
#define L 5000000
using namespace std;

long long num, m;
long long a[L];
long long size[L];
long long lazy[L];
long long sum[L];
long long n, v, l, r, j, k;
char bj;

void build(long long n, long long l, long long r){
    size[n] = r - l + 1;
    if (l == r){
        sum[n] = a[l];
        return ;
    }
    long long mid = (l + r) >> 1;
    build(ls, l, mid), build(rs, mid + 1, r);
    sum[n] = sum[ls] + sum[rs];
}

void down(long long n){
    sum[ls] += lazy[n] * size[ls];
    sum[rs] += lazy[n] * size[rs];
    lazy[ls] += lazy[n], lazy[rs] += lazy[n], lazy[n] = 0;
}

void add(long long n, long long l, long long r, long long xl, long long xr, long long v){
    if (l != r) down(n);
    if (xl == l && r == xr){
        sum[n] += v * size[n];
        lazy[n] += v;
        return ;
    }
    long long mid = (l + r) >> 1;
    if (xr <= mid) add(ls, l, mid, xl, xr, v);
    else if (xl > mid) add(rs, mid + 1, r, xl, xr, v);
    else add(ls, l, mid, xl, mid, v), add(rs, mid + 1, r, mid + 1, xr, v);
    sum[n] = sum[ls] + sum[rs];
}

long long query(long long n, long long l, long long r, long long xl, long long xr){
    if (l != r) down(n);
    if (xl <= l && r <= xr) return sum[n];
    long long mid = (l + r) >> 1;
    if (xr <= mid) return query(ls, l, mid, xl, xr);
    else if (xl > mid) return query(rs, mid + 1, r, xl, xr);
    else return query(ls, l, mid, xl, mid) + query(rs, mid + 1, r, mid + 1, xr);
}

int main()
{
    scanf("%d %d", &num, &m);
    for (int i = 0; i < num; ++i)    scanf("%d", &a[i]);
    build(1, 0, num - 1);
    for (int i = 1; i <= m; ++i){
        cin >> bj;
        if (bj == 'C') scanf("%lld %lld %lld", &l, &r, &v), add(1, 0, num - 1, l, r, v);
        if (bj == 'Q') scanf("%lld %lld", &j, &k), printf("%lld\n", query(1, 0, num - 1, j, k));
    }
    return 0;
}

CJOJ P2008 - 【JSOI2008】最大数

线段树求最大值的板子题

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define LL long long
#define ls (n << 1)
#define rs (n << 1 | 1)
#define L 1000000
using namespace std;

LL t, l, x;
LL m, d;
char bj;
LL num[L];

void buildtree(int n, int l, int r) {
    if (l == r) {
        num[n] = 0;
        return ;
    }
    int mid = (l + r) >> 1;
    buildtree(ls, l, mid), buildtree(rs, mid + 1, r);
    num[n] = max(num[ls], num[rs]);
}

LL query(int n, int l, int r, int a, int b) {
    if (a == l && r == b) return num[n];
    int mid = (l + r) >> 1;
    if (b <= mid) return query(ls, l, mid, a, b);
    else if (mid < a) return query(rs, mid + 1, r, a, b);
    else return max(query(ls, l, mid, a, mid), query(rs, mid + 1, r, mid + 1, b));
}

void add(int n, int l, int r, int a, LL v) {
    if (l == a && r == a) {
        num[n] = v;
        return ;
    }
    int mid = (l + r) >> 1;
    if (a <= mid) add(ls, l, mid, a, v);
    else add(rs, mid + 1, r, a, v);
    num[n] = max(num[ls], num[rs]);
}

int main() {
    scanf("%lld %lld", &m, &d);
    buildtree(1, 1, 210000);

    for (int i = 1; i <= m; ++i) {
        cin >> bj >> x;
        if (bj == 'Q') {
            t = query(1, 1, m, l - x + 1, l);
            printf("%lld", t);
        }
        if (bj == 'A') add(1, 1, m, ++l, (t + x) % d);
    }
    return 0;
}

CJOJ P2265 线段覆盖

线段树区间修改板子题

#include <iostream>
#include <cstdio>
#include <cstdlib>
#define L 1000000
#define ls (n << 1)
#define rs (n << 1 | 1)
using namespace std;

int l, n;
int m, a, t;
int x[L], cnt[L], len[L], rc[L], lc[L], lazy[L];

void down(int n, int l, int r) {
    if (lazy[n] > 0) {
        cnt[n] = rc[n] = lc[n] = 1;
        len[n] = r - l + 1;
    }
    else {
        lc[n] = lc[ls];
        rc[n] = rc[rs];
        cnt[n] = cnt[ls] + cnt[rs];
        if (rc[ls] == 1 && lc[rs] == 1) cnt[n]--;
        len[n] = len[ls] + len[rs];
    }
}

void add(int n, int l, int r, int a, int b, int v) {
    if (a <= l && r <= b) {
        lazy[n] += v;
        if (l == r) {
            if (lazy[n] > 0)
                lazy[n] = rc[n] = lc[n] = cnt[n] = len[n] = 1;
            else lazy[n] = rc[n] = lc[n] = cnt[n] = len[n] = 0;
        }
        else down(n, l, r);
    }
    else {
        int mid = (l + r) >> 1;
        if (b > mid) add(rs, mid + 1, r, a, b, v);
        if (a <= mid) add(ls, l, mid, a, b, v);
        down(n, l, r);
    }
}

int main() {
    scanf("%d %d", &l, &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d %d %d", &m, &a, &t);
        if (m == 1) add(1, 1, l, a, a + t - 1, 1);
        else add(1, 1, l, a, a + t - 1, -1);
        printf("%d %d\n", cnt[1], len[1]);
    }
    return 0;
}

总结未做的题
CJOJ P1810 - 【USACO 5.5.1】矩形周长

——CYCKN