243. 一个简单的整数问题2(线段树,懒标记,树状数组,《算法竞赛进阶指南》)
https://www.acwing.com/problem/content/244/
给定一个长度为 N 的数列 A,以及 M 条指令,每条指令可能是以下两种之一:
C l r d
,表示把 A[l],A[l+1],…,A[r] 都加上 d。Q l r
,表示询问数列中第 l∼r 个数的和。
对于每个询问,输出一个整数表示答案。
输入格式
第一行两个整数 N,M。
第二行 N 个整数 A[i]。
接下来 M 行表示 M 条指令,每条指令的格式如题目描述所示。
输出格式
对于每个询问,输出一个整数表示答案。
每个答案占一行。
数据范围
1≤N,M≤105
|d|≤10000
|A[i]|≤109
输入样例:
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
输出样例:
4
55
9
15
解析:
线段数做法(懒标记):
进行区间操作时如果当前节点包含在区间内就直接更新,否则递归到子节点
区间修改需要用到懒标记,就是这段区间要加几先记录下来,要用到区间内部的信息时再下推到子节点
带懒标记的线段树可以在O(logn)的时间复杂度内完成各种区间操作
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>
#include<sstream>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = 0x3f3f3f3f;
const LL Mod = 1e9;
const int N = 1e5 + 10, M = 250 + 10, P = 110;
int n, m;
LL w[N];
struct Node {int l, r;LL sum, add;
}tr[N * 4];void pushup(int u) {tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
}void pushdown(int u) {Node& root = tr[u], & l = tr[u << 1], & r = tr[u << 1 | 1];if (root.add) {l.sum +=(LL) (l.r - l.l + 1) * root.add, l.add += root.add;r.sum += (LL)(r.r - r.l + 1) * root.add, r.add += root.add;root.add = 0;}
}void build(int u, int l, int r) {if (l == r) {tr[u] = { l,l,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);pushup(u);}
}void modify(int u, int l, int r, int d) {if (tr[u].l >= l && tr[u].r <= r) {tr[u].sum += (tr[u].r - tr[u].l + 1) * d;tr[u].add += d;}else {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);pushup(u);}
}LL query(int u, int l, int r) {if (tr[u].l >= l && tr[u].r <= r) {return tr[u].sum;}else {pushdown(u);int mid = tr[u].l + tr[u].r >> 1;LL ret = 0;if (l <= mid)ret += query(u << 1, l, r);if (r > mid)ret += query(u << 1 | 1, l, r);pushup(u);return ret;}
}int main() {cin >> n >> m;for (int i = 1; i <= n; i++) {scanf("%lld", &w[i]);}build(1, 1, n);char op[2];int l, r, d;while (m--) {scanf("%s", op);if (op[0] == 'C') {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;
}
树状数组:
算前x个数的和时,差分后的第i(0<i<=x)个数加了x-i+1次
再定义一个数组B,B[i]=(A[i]-A[i-1])*i,c2为B的树状数组
求前缀和时返回c1[i]*(x+1)-c2[i]
神奇的事情出现了:
c1[i]*(x+1)中i往前lowbit(i)个数都被加了x+1次,
c2[i]中i往前lowbit(i)个数分别被加了下标次,
下标为j的数正好会被加x-j+1次
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>
#include<sstream>
#include<deque>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = 0x3f3f3f3f;
const LL Mod = 1e9;
const int N = 1e5 + 10, M = 250 + 10, P = 110;
int n, m;
int a[N];
LL tr1[N], tr2[N];int lowbit(int x) {return x & -x;
}void add(LL tr[], int x, LL d) {for (int i = x; i <= n; i += lowbit(i))tr[i] += d;
}LL sum(LL tr[], int x) {LL ret = 0;for (int i = x; i; i -= lowbit(i))ret += tr[i];return ret;
}LL pre_sum(int x) {return sum(tr1, x) * (x + 1) - sum(tr2, x);
}int main() {cin >> n >> m;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 1; i <= n; i++) {add(tr1, i,(LL) a[i] - a[i - 1]);add(tr2, i, (LL)i * (a[i] - a[i - 1]));}while (m--) {char op[2];scanf("%s", op);if (op[0] == 'C') {int l, r, d;scanf("%d%d%d", &l, &r, &d);add(tr1, l, (LL) d),add(tr1,r+1,(LL)(-d));add(tr2, l, (LL) l* d),add(tr2, r + 1, (LL)(r + 1) *(-d));}else {int l, r;scanf("%d%d", &l, &r);printf("%lld\n", pre_sum(r)-pre_sum(l-1));}}return 0;
}