created by Dejavu
(不斷更新中)
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概念
線段樹，類似區間樹，是一個完全二叉樹，它在各個節點保存一條線段（數組中的一段子數組），主要用于高效解決連續區間的動態查詢問題，由于二叉結構的特性，它基本能保持每個操作的復雜度為O(lgN)，你可以在數據結構可視(visualgo)網站中看到算法的具體搭建過程。
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c++ 模板代碼
這里的代碼是對區間寫入優化后的代碼，加入了lazy_tag用來標記區間更新的位置
//最好不要用頭文件bits/stdc++.h 因為一些編譯器會不支持這個頭文件
#include <iostream>
using namespace std;
#define MaxN 50005
class SegTree {
public:
int left, right;
int sum;
int lazy_tag;
int mid() { return (left + right) >> 1; }
};
SegTree tree[MaxN];
void pushUp(int rt) { tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum; }
void pushDown(int rt, int len) {
int tag = tree[rt].lazy_tag;
if (tag) {
tree[rt << 1].lazy_tag += tag;
tree[rt << 1 | 1].lazy_tag += tag;
tree[rt << 1].sum += tag*(len - (len >> 1));
tree[rt << 1 | 1].sum += tag*(len >> 1);
tree[rt].lazy_tag = 0;
}
}
void build(int rt, int left, int right) {
tree[rt].left = left;
tree[rt].right = right;
tree[rt].lazy_tag = 0;
if (left == right) { cin >> tree[rt].sum;return; }
int mid = tree[rt].mid();
build(rt << 1, left, mid);
build(rt << 1 | 1, mid + 1, right);
pushUp(rt);
}
void update(int rt, int val, int left, int right) {
if (tree[rt].left == left && tree[rt].right == right) {
tree[rt].lazy_tag += val;
tree[rt].sum += val*(right - left + 1);
return;
}
if (tree[rt].left == tree[rt].right) return;
pushDown(rt, tree[rt].right - tree[rt].left + 1);
int mid = tree[rt].mid();
if (right <= mid) update(rt << 1, val, left, right);
else if (left > mid) update(rt << 1 | 1, val, left, right);
else {
update(rt << 1, val, left, mid);
update(rt << 1 | 1, val, mid + 1, right);
}
pushUp(rt);
}
int query(int rt, int left, int right) {
if (tree[rt].left == left && tree[rt].right == right) return tree[rt].sum;
pushDown(rt, tree[rt].right - tree[rt].left + 1);
int mid = tree[rt].mid();
int sum(0);
if (right <= mid) sum += query(rt << 1, left, right);
else if (left > mid) sum += query(rt << 1 | 1, left, right);
else {
sum += query(rt << 1, left, mid);
sum += query(rt << 1 | 1, mid + 1, right);
}
return sum;
}
int main() {
ios::sync_with_stdio(false);
int n;cin >> n;
build(1, 1, n);
char cmd;
int in1, in2, val;
while (cin >> cmd, cmd != 'E') {
if (cmd == 'Q') { cin >> in1 >> in2;cout << query(1, in1, in2) << endl; }
if (cmd == 'A') { cin >> in1 >> in2 >> val;update(1, val, in1, in2); }
}
}
例題
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單點更新
最最基礎的線段樹,只更新葉子節點,然后把信息用pushUP(int rt)這個函數更新上來 - hdu1166 敵兵布陣
- hdu1754 I Hate It
- hdu1394 Minimum Inversion Number
- hdu2795 Billboard
- poj2828 Buy Tickets
- poj2886 Who Gets the Most Candies?
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成段更新
需要用到延遲標記(或者說懶惰標記),簡單來說就是每次更新的時候不要更新到底,用延遲標記使得更新延遲到下次需要更新or詢問到的時候 - hdu1698 Just a Hook
- poj3468 A Simple Problem with Integers
- poj2528 Mayor’s posters
- poj3225 Help with Intervals
- poj1436 Horizontally Visible Segments
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區間合并
這類題目會詢問區間中滿足條件的連續最長區間,所以pushUp的時候需要對左右兒子的區間進行合并 - poj2991 Crane
- hdu3308 LCIS
- hdu3397 Sequence operation
- hdu2871 Memory Control
- hdu1540 Tunnel Warfare
- CF46-D Parking Lot
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掃描線
這類題目需要將一些操作排序,然后從左到右用一根掃描線(當然是在我們腦子里)掃過去
最典型的就是矩形面積并,周長并等題 - hdu1542 Atlantis
- hdu1828 Picture
- hdu3265 Posters
- hdu3642 Get The Treasury
- poj2482 Stars in Your Window
- poj2464 Brownie Points II
- hdu3255 Farming
- ural1707 Hypnotoad’s Secret
- uva11983 Weird Advertisement
題解和代碼性能分析
- 單點更新
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hdu1166 敵兵布陣
求區間和
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hdu1166 敵兵布陣
//c++ 用類似c的方法寫
#include <stdio.h>
#include <string.h>
#define MaxN 50000
int t[MaxN * 4];
inline void pushUp(int rt) { t[rt] = t[rt << 1] + t[rt << 1 | 1]; }
void build(int rt,int l,int r) {
if (l == r) {scanf("%d",&t[rt]);return;}
int m = (l + r) >> 1;
build(rt << 1, l, m);
build(rt << 1 | 1, m + 1, r);
pushUp(rt);
}
void update(int rt, int l, int r, int p, int v) {
if (l == r) { t[rt] += v;return; }
int m = (l + r) >> 1;
if (p <= m) update(rt << 1, l, m, p, v);
else update(rt << 1 | 1, m + 1, r, p, v);
pushUp(rt);
}
int query(int rt,int l,int r, int _l, int _r) {
if (_l <= l && r <= _r) return t[rt];
int m = (l + r) >> 1;
int sum(0);
if (_r <= m) sum += query(rt << 1, l, m, _l, _r);
else if (_l > m) sum += query(rt << 1 | 1, m + 1, r, _l, _r);
else {
sum += query(rt << 1, l, m, _l, _r);
sum += query(rt << 1 | 1, m + 1, r, _l, _r);
}
return sum;
}
void main() {
int n, tn;scanf("%d",&n);tn = n;
while (tn--) {
int num;scanf("%d", &num);
build(1, 1, num);
printf("Case %d:\n", n - tn);
char cmd[10];
int in1, in2;
while (scanf("%s", &cmd)) {
if (cmd[0] == 'E') break;
scanf("%d%d", &in1, &in2);
if (cmd[0] == 'A') update(1, 1, num, in1, in2);
else if (cmd[0] == 'S') update(1, 1, num, in1, -in2);
else printf("%d\n", query(1, 1, num, in1, in2));
memset(cmd, '\0', sizeof(cmd));
}
}
}
上面代碼在hdu運行的結果
//更改query函數
int sum(0);
void query(int rt, int l, int r, int _l, int _r) {
if (_l <= l && r <= _r) {sum += t[rt];return;}
int m = (l + r) >> 1;
if (_r <= m) query(rt << 1, l, m, _l, _r);
else if (_l > m) query(rt << 1 | 1, m + 1, r, _l, _r);
else {
query(rt << 1, l, m, _l, _r);
query(rt << 1 | 1, m + 1, r, _l, _r);
}
}
更改query函數將sum作為全局變量
//c++ 用c寫法下，全宏定義
#include <stdio.h>
#include <string.h>
#define MaxN 50000
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define mid int m=(l+r)>>1
#define base int rt,int l,int r
#define we_base 1,1,num
int t[MaxN*4];
inline void pushUp(int rt) { t[rt] = t[rt << 1] + t[rt << 1 | 1]; }
void build(base) {
if (l == r) { scanf("%d", &t[rt]);return; }
mid;
build(lson);
build(rson);
pushUp(rt);
}
void update(base, int p, int v) {
if (l == r) { t[rt] += v;return; }
mid;
if (p <= m) update(lson, p, v);
else update(rson, p, v);
pushUp(rt);
}
int query(base, int _l, int _r) {
if (_l <= l && r <= _r) return t[rt];
mid;
int sum(0);
if (_r <= m) sum += query(lson, _l, _r);
else if (_l > m) sum += query(rson, _l, _r);
else {
sum += query(lson, _l, _r);
sum += query(rson, _l, _r);
}
return sum;
}
void main() {
int n, tn;scanf("%d", &n);tn = n;
while (tn--) {
int num;scanf("%d", &num);
build(we_base);
printf("Case %d:\n", n - tn);
char cmd[10];
int in1, in2;
while (scanf("%s", &cmd)) {
if (cmd[0] == 'E') break;
scanf("%d%d", &in1, &in2);
if (cmd[0] == 'A') update(we_base, in1, in2);
else if (cmd[0] == 'S') update(we_base, in1, -in2);
else printf("%d\n", query(we_base, in1, in2));
memset(cmd, '\0', sizeof(cmd));
}
}
}
c寫法全宏定義
可見使用宏定義的算法運行速度比普通寫法稍快一些，而且寫代碼的速度將會比普通寫法快
//用c++的形式寫
#include <iostream>
#include <string>
using namespace std;
#define MaxN 50000
int t[MaxN * 4];
inline void pushUp(int rt) { t[rt] = t[rt << 1] + t[rt << 1 | 1]; }
void build(int rt,int l,int r) {
if (l == r) { cin >> t[rt];return; }
int m = (l + r) >> 1;
build(rt << 1, l, m);
build(rt << 1 | 1, m + 1, r);
pushUp(rt);
}
void update(int rt, int l, int r, int p, int v) {
if (l == r) { t[rt] += v;return; }
int m = (l + r) >> 1;
if (p <= m) update(rt << 1, l, m, p, v);
else update(rt << 1 | 1, m + 1, r, p, v);
pushUp(rt);
}
int query(int rt,int l,int r, int _l, int _r) {
if (_l <= l && r <= _r) return t[rt];
int m = (l + r) >> 1;
int sum(0);
if (_r <= m) sum += query(rt << 1, l, m, _l, _r);
else if (_l > m) sum += query(rt << 1 | 1, m + 1, r, _l, _r);
else {
sum += query(rt << 1, l, m, _l, _r);
sum += query(rt << 1 | 1, m + 1, r, _l, _r);
}
return sum;
}
void main() {
ios::sync_with_stdio(false);
int n, tn;cin >> n;tn = n;
while (tn--) {
int num;cin >> num;
build(1, 1, num);
cout << "Case " << n - tn <<":" << endl;
string cmd;
int in1, in2;
while (cin >> cmd, cmd != "End") {
cin>> in1 >> in2;
if (cmd == "Add") update(1, 1, num, in1, in2);
else if (cmd == "Sub") update(1, 1, num, in1, -in2);
else cout << query(1, 1, num, in1, in2) << endl;
}
}
}
ios::sync_with_stdio(false)
ios::sync_with_stdio(true)
這里的1000ms并不是真正的運行時間而是超時后就停止運行了，就是說當設置為true時，運行時間是遠遠超過設置為false時的時間的，sync_with_stdio(bool)這個語句是詢問是否與stdio庫同步，同步過程會耗費很多時間
//c++ 下全宏定義寫法
#include <iostream>
#include <string>
using namespace std;
#define MaxN 50000
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define mid int m=(l+r)>>1
#define base int rt,int l,int r
#define we_base 1,1,num
int t[MaxN*4];
inline void pushUp(int rt) { t[rt] = t[rt << 1] + t[rt << 1 | 1]; }
void build(base) {
if (l == r) { cin >> t[rt];return; }
mid;
build(lson);
build(rson);
pushUp(rt);
}
void update(base, int p, int v) {
if (l == r) { t[rt] += v;return; }
mid;
if (p <= m) update(lson, p, v);
else update(rson, p, v);
pushUp(rt);
}
int query(base, int _l, int _r) {
if (_l <= l && r <= _r) return t[rt];
mid;
int sum(0);
if (_r <= m) sum += query(lson, _l, _r);
else if (_l > m) sum += query(rson, _l, _r);
else {
sum += query(lson, _l, _r);
sum += query(rson, _l, _r);
}
return sum;
}
void main() {
ios::sync_with_stdio(false);
int n, tn;cin >> n;tn = n;
while (tn--) {
int num;cin >> num;
build(we_base);
cout << "Case " << n - tn << ":" << endl;
string cmd;
int in1, in2;
while (cin >> cmd, cmd != "End") {
cin >> in1 >> in2;
if (cmd == "Add") update(we_base, in1, in2);
else if (cmd == "Sub") update(we_base, in1, -in2);
else cout << query(we_base, in1, in2) << endl;
}
}
}
c++ 下全宏定義寫法
然而使用c++下的頭文件和空間名，似乎用宏定義效果并不好
代碼運行各項性能對比
c寫法
c++下普通c的寫法
更改query函數將sum作為全局變量
c寫法全宏定義
c++寫法
ios::sync_with_stdio(false)
ios::sync_with_stdio(true)
c++ 下全宏定義寫法
用c++的輸入輸出流在性能上會比c的輸入輸出流慢一些，即使加了去同步語句，所以acm的代碼最好采用c的輸入輸出流
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hdu1754 I Hate It
求區間最大值
#include <cstdio>
#define s scanf
#define p printf
#define base int rt,int l,int r
#define we_base 1,1,num
#define max(v1,v2) v1>v2?v1:v2
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define mid int m=(l+r)>>1
#define MaxN 200000
int t[MaxN * 4];
inline void pushUp(int rt) { t[rt] = max(t[rt << 1], t[rt << 1 | 1]); }
void build(base) {
if (l == r) s("%d", &t[rt]);
else {
mid;
build(lson);
build(rson);
pushUp(rt);
}
}
void update(base, int pos, int v) {
if (l == r) t[rt] = v;
else {
mid;
if (pos <= m) update(lson, pos, v);
else update(rson, pos, v);
pushUp(rt);
}
}
int query(base, int _l, int _r) {
if (_l <= l&&_r >= r) return t[rt];
mid;
if (_r <= m) return query(lson, _l, _r);
else if (_l > m) return query(rson, _l, _r);
else {
int t1 = query(lson, _l, m);
int t2 = query(rson, m + 1, _r);
return max(t1, t2);
}
}
}
void main() {
int num, m;
while (s("%d%d", &num, &m) != EOF) {
build(we_base);
char cmd;
int in1, in2;
while (m--) {
s("\n%c%d%d", &cmd, &in1, &in2);
if (cmd == 'U') update(we_base, in1, in2);
else p("%d\n", query(we_base, in1, in2));
}
}
}
