一、線段樹
Reference:https://www.cnblogs.com/AC-King/p/7789013.html
要解決的問題:
1.查詢區間[L,R]之間的最值
2.修改a[i]爲x;
明確可以解決的問題:
必須是滿足區間可加性的問題,例如:
符合區間加法的例子:
數字之和——總數字之和 = 左區間數字之和 + 右區間數字之和
最大公因數(GCD)——總GCD = gcd( 左區間GCD , 右區間GCD );
最大值——總最大值=max(左區間最大值,右區間最大值)
不符合區間加法的例子:
衆數——只知道左右區間的衆數,沒法求總區間的衆數
01序列的最長連續零——只知道左右區間的最長連續零,沒法知道總的最長連續零
線段樹的建立
#include <iostream>
using namespace std;
const int maxn = 110;
int a[maxn];
int minv[4 * maxn];
void pushup(int id) {
minv[id] = min(minv[id << 1],minv[id << 1 | 1]);
}
void build(int id,int l,int r) {
if(l == r) {
minv[id] = a[l];
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
pushup(id);
}
int main() {
int n;
cin >> n;
for(int i = 1;i <= n; ++i) {
cin >> a[i];
}
build(1,1,n);
return 0;
}
2.線段樹的單點更新
#include <iostream>
using namespace std;
const int maxn = 110;
int a[maxn];
int minv[4 * maxn];
void pushup(int id) {
minv[id] = min(minv[id << 1], minv[id << 1 | 1]);
}
void build(int id, int l, int r) {
if (l == r) {
minv[id] = a[l];
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
pushup(id);
}
void update(int id,int l,int r,int x,int v) {
if(l == r) {
minv[id] = v;
return;
}
int mid = (l + r) >> 1;
if(x <= mid) {
update(id << 1, l, mid, x, v);
}else {
update(id << 1 | 1, mid + 1, r, x ,v);
}
pushup(id);
}
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
build(1, 1, n);
int q;
cin >> q;
for(int i = 0;i < q; ++i) {
int x,v;
cin >> x >> v;
update(1, 1, n, x, v);
}
return 0;
}
3.線段樹的區間查詢
#include <iostream>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 110;
int a[maxn];
int minv[4 * maxn];
void pushup(int id) {
minv[id] = min(minv[id << 1], minv[id << 1 | 1]);
}
void build(int id, int l, int r) {
if (l == r) {
minv[id] = a[l];
return;
}
int mid = (l + r) >> 1;
build(id << 1, l, mid);
build(id << 1 | 1, mid + 1, r);
pushup(id);
}
void update(int id, int l, int r, int x, int v) {
if (l == r) {
minv[id] = v;
return;
}
int mid = (l + r) >> 1;
if (x <= mid) {
update(id << 1, l, mid, x, v);
} else {
update(id << 1 | 1, mid + 1, r, x, v);
}
pushup(id);
}
int query(int id, int l, int r, int x, int y) { // 區間查詢
if(x <= l && r <= y) {
return minv[id];
}
int mid = (l + r) >> 1;
int ans = inf;
if(x <= mid) {
ans = min(ans,query(id << 1,l, mid, x, y));
}
if(y > mid) {
ans = min(ans, query(id << 1 | 1,mid + 1, r, x, y));
}
return ans;
}
int query(int id, int l, int r, int x) { //單點查詢
if (l == r) {
return minv[id];
}
int mid = (l + r) >> 1;
if (mid <= x) {
return query(id << 1, l, mid, x);
} else {
return query(id << 1 | 1, mid + 1, r, x);
}
}
int main() {
int n;
cin >> n;
for (int i = 1; i <= n; ++i) {
cin >> a[i];
}
build(1, 1, n);
int q;
cin >> q;
for (int i = 0; i < q; ++i) {
int x, v;
cin >> x >> v;
update(1, 1, n, x, v);
}
int p;
cin >> p;
for(int i = 0;i < p; ++i){
int l, r;
cin >> l >> r;
cout << query(1,1,n,l,r) << endl;
}
return 0;
}
二、樹狀數組
1.求lowbit
lowboy(x) = x & (-x);
2.查詢及更新
#include <iostream>
using namespace std;
const int maxn = 110;
int C[maxn], n;
int lowbit(int x) {
return x &(-x);
}
int getsum(int x) {
int res = 0;
for(int i = x;i > 0; i-=lowbit(i)) {
res += C[i];
}
return res;
}
void update(int x,int v) {
for(int i = x; i <= n; i+= lowbit(i)) {
C[i] += v;
}
}
int main() {
cin >> n;
for(int i = 1; i <= n; ++i) {
int v;
cin >> v;
update(i,v);
}
int q;
cin >> q;
while(q--) {
int l,r;
cin >> l>> r;
cout << getsum(r) - getsum(l - 1) << endl;
}
return 0;
}