线段树

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);}

• 下面看几道例题（题目差不多）
  Codevs 1080线段树练习

#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 500000using 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 1000000using 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 5000000using 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 1000000using 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