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POJ-3468 A Simple Problem with Integers
题意: 给定一个数字序列, 有两张操作: 1. 查询[L, R] 的所有数字之和。2. 给定区间[L, R] 的所有数加C.
分析: 很明显的线段树区间更新问题, 但是在这里要引入一种新的算法------分块。 分块的思想就是把区间分为相等的块, 大小一般为sqrt(n), 在进行操作的时候直接对块进行操作, 查询和更新的时间复杂度可以达到sqrt(n), 与线段树的时间复杂度不相上下。 这道题是一道很经典的分块和线段树区间更新问题, 下面的代码可以当做模板,下面给出两种代码, 即线段树和分块。要注意的是, 在分块的时候也要使用lazy标记, 表示在分块区间的累加值, 在查询的时候直接加入到结果中即可。
分块代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1e5 + 7;
int block, num, l[MAXN], r[MAXN], belong[MAXN];
int n, q;
long long a[MAXN], sum[MAXN], lazy[MAXN];
void build()
{
memset(sum, 0, sizeof(sum));
block = sqrt(n);
num = n / block;
if (n % block)
num++;
for (int i = 1; i <= num; i++)
{
l[i] = (i - 1) * block + 1;
r[i] = i * block;
}
r[num] = n;
for (int i = 1; i <= n; i++)
belong[i] = (i - 1) / block + 1;
for (int i = 1; i <= num; i++)
{
for (int j = l[i]; j <= r[i]; j++)
sum[i] += a[j];
}
}
void debug()
{
for (int i = 1; i <= num; i++)
cout << sum[i] << " ";
cout << endl;
for (int i = 1; i <= num; i++)
cout << lazy[i] << " ";
cout << endl;
}
void update(int x, int y, long long z)
{
if (belong[x] == belong[y])
{
for (int i = x; i <= y; i++)
{
a[i] += z;
sum[belong[x]] += z;
}
return;
}
for (int i = x; i <= r[belong[x]]; i++)
{
a[i] += z;
sum[belong[x]] += z;
}
for (int i = belong[x] + 1; i < belong[y]; i++)
{
lazy[i] += z;
}
for (int i = l[belong[y]]; i <= y; i++)
{
a[i] += z;
sum[belong[y]] += z;
}
}
long long ask(int x, int y)
{
long long res = 0;
if (belong[x] == belong[y])
{
for (int i = x; i <= y; i++)
{
res += a[i] + lazy[belong[x]];
}
return res;
}
for (int i = x; i <= r[belong[x]]; i++)
{
res += a[i] + lazy[belong[x]];
}
for (int i = belong[x] + 1; i < belong[y]; i++)
{
res += sum[i] + 1l * lazy[i] * block;
}
for (int i = l[belong[y]]; i <= y; i++)
{
res += a[i] + lazy[belong[y]];
}
return res;
}
int main()
{
while (scanf("%d%d", &n, &q) != EOF)
{
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]);
build();
memset(lazy, 0, sizeof(lazy));
for (int i = 0; i < q; i++)
{
char op[10];
int x, y;
long long z;
scanf("%s", op);
//debug();
if (op[0] == 'Q')
{
scanf("%d%d", &x, &y);
printf("%lld\n", ask(x, y));
}
else
{
scanf("%d%d%lld", &x, &y, &z);
update(x, y, z);
}
}
}
}
线段树代码:
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
struct _Node {
int l, r;
ll sum;
int mid() { return (l + r) / 2; }
};
const int MAXN = 100005;
_Node tree[MAXN << 2];
ll lazy[MAXN << 2];
ll value[MAXN];
void push_up(int root) {
tree[root].sum = tree[root << 1].sum + tree[root << 1 | 1].sum;
}
void push_down(int root, int len) { //更新lazy
if (lazy[root]) {
lazy[root << 1] += lazy[root];
lazy[root << 1 | 1] += lazy[root];
tree[root << 1].sum += lazy[root] * (len - len / 2);
tree[root << 1 | 1].sum += lazy[root] * (len / 2);
lazy[root] = 0;
}
}
void init(int root, int l, int r) {
tree[root].l = l;
tree[root].r = r;
if (l == r) {
tree[root].sum = value[l];
return;
}
init(root << 1, l, (l + r) / 2);
init(root << 1 | 1, (l + r) / 2 + 1, r);
push_up(root);
}
void update(int root, int l, int r, int value) {
if (tree[root].l == l && tree[root].r == r) {
lazy[root] += value;
tree[root].sum += (ll)(r - l + 1) * value;
return;
}
if (tree[root].l == tree[root].r) {
return;
}
int mid = tree[root].mid();
push_down(root, tree[root].r - tree[root].l + 1);
if (l > mid)
update(root << 1 | 1, l, r, value);
else if (r <= mid)
update(root << 1, l, r, value);
else {
update(root << 1, l, mid, value);
update(root << 1 | 1, mid + 1, r, value);
}
push_up(root);
}
ll query(int root, int l, int r) {
if (tree[root].l == l && tree[root].r == r) {
return tree[root].sum;
}
int mid = tree[root].mid();
push_down(root, tree[root].r - tree[root].l + 1);
if (l > mid) {
return query(root << 1 | 1, l, r);
} else if (r <= mid) {
return query(root << 1, l, r);
} else {
return query(root << 1, l, mid) + query(root << 1 | 1, mid + 1, r);
}
}
int main() {
int n, m, a, b, c;
char op[10];
while (scanf("%d%d", &n, &m) != EOF) {
memset(lazy, 0, sizeof(lazy));
for (int i = 1; i <= n; ++i)
scanf("%lld", &value[i]);
init(1, 1, n);
for (int i = 0; i < m; ++i) {
scanf("%s", op);
scanf("%d%d", &a, &b);
if (op[0] == 'Q') {
if (a > b)
swap(a, b);
printf("%lld\n", query(1, a, b));
} else {
scanf("%d", &c);
update(1, a, b, c);
}
}
}
return 0;
}