LOJ6284 数列分块入门 8
标签
- 分块入门
前言
- 我的csdn和博客园是同步的,欢迎来访danzh-博客园~
简明题意
- 维护序列,支持两种操作:
- 查询区间中某种权值的出现次数
- 将整个区间[l,r]全部改为c
思路
- 这个暴力得想不到呀。
- 真的很暴力,实际上开一个数组去维护一下每一块的值是否是一样的,再开一个数组记录如果某一块的值相同那么这个值是多少。然后就很简单了惹
注意事项
- 无
总结
- 无
AC代码
#pragma GCC optimize(2)
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e5 + 10;
int read()
{
int x = 0, f = 1; char ch = getchar();
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0'&&ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
int n, a[maxn];
int pos[maxn], len, val[maxn];
bool is_zh[maxn];
int cal(int k, int c)
{
if (is_zh[k])
return val[k] == c ? len : 0;
int ans = 0;
for (int i = k * len - len + 1; i <= k * len; i++)
if (a[i] == c)
ans++;
return ans;
}
void reset(int k)
{
if (is_zh[k])
{
for (int i = k * len - len + 1; i <= min(k * len, n); i++)
a[i] = val[k];
is_zh[k] = 0;
}
}
int deal(int l, int r, int c)
{
//查询
int ans = 0;
reset(pos[l]);
for (int i = l; i <= min(pos[l] * len, r); i++)
if (a[i] == c)
ans++;
if (pos[l] != pos[r])
{
reset(pos[r]);
for (int i = pos[r] * len - len + 1; i <= r; i++)
if (a[i] == c)
ans++;
}
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
ans += cal(i, c);
//修改
for (int i = l; i <= min(pos[l] * len, r); i++)
a[i] = c;
if (pos[l] != pos[r])
{
for (int i = pos[r] * len - len + 1; i <= r; i++)
a[i] = c;
}
for (int i = pos[l] + 1; i <= pos[r] - 1; i++)
is_zh[i] = 1, val[i] = c;
return ans;
}
void solve() {
scanf("%d", &n);
len = sqrt(n);
for (int i = 1; i <= n; i++)
a[i] = read(), pos[i] = (i - 1) / len + 1;
for (int i = 1; i <= n; i++) {
int l, r, c;
l = read(), r = read(), c = read();
printf("%d\n", deal(l, r, c));
}
}
int main() {
freopen("Testin.txt", "r", stdin);
freopen("Testout.txt", "w", stdout);
solve();
return 0;
}