Description
给定一个长为(n)的区间,(m)次操作,给子区间涂一种颜色,或者查询一个区间内有多少种颜色。颜色种类数(le30),初始颜色为(1),(n,mle10^5)。
Solution
开始想的是开(30)棵线段树,或者每个区间开一个(vector)存颜色,但后来才发现其实完全没必要。注意到颜色种类数很少,所以其实可以把它压成一个状态(S),其中(S)从左往右数第(i)位是(1)表示这个区间内有(i)这种颜色。
那么这就很好维护了。(pushdown)就直接覆盖,(pushup)就是对儿子进行或运算就好了。
Code
#include <bits/stdc++.h>
using namespace std;
#define ls(x) (x << 1)
#define rs(x) (x << 1 | 1)
int n, m, mx;
struct node
{
int l, r, col, tag;
}t[400005];
char ch[2];
int read()
{
int x = 0, fl = 1; char ch = getchar();
while (ch < '0' || ch > '9') { if (ch == '-') fl = -1; ch = getchar();}
while (ch >= '0' && ch <= '9') {x = (x << 1) + (x << 3) + ch - '0'; ch = getchar();}
return x * fl;
}
void push_up(int p)
{
t[p].col = (t[ls(p)].col | t[rs(p)].col);
return;
}
void push_down(int p)
{
if (!t[p].tag) return;
t[ls(p)].tag = t[p].tag;
t[rs(p)].tag = t[p].tag;
t[ls(p)].col = (1 << (t[p].tag - 1));
t[rs(p)].col = (1 << (t[p].tag - 1));
t[p].tag = 0;
return;
}
void build(int p, int l0, int r0)
{
t[p].l = l0; t[p].r = r0;
if (l0 == r0)
{
t[p].col = 1;
return;
}
int mid = (t[p].l + t[p].r) >> 1;
build(ls(p), l0, mid);
build(rs(p), mid + 1, r0);
push_up(p);
return;
}
void update(int p, int l0, int r0, int d)
{
if (l0 <= t[p].l && t[p].r <= r0)
{
t[p].col = (1 << (d - 1));
t[p].tag = d;
return;
}
push_down(p);
int mid = (t[p].l + t[p].r) >> 1;
if (l0 <= mid) update(ls(p), l0, r0, d);
if (r0 > mid) update(rs(p), l0, r0, d);
push_up(p);
return;
}
int query(int p, int l0, int r0)
{
if (l0 <= t[p].l && t[p].r <= r0) return t[p].col;
push_down(p);
int mid = (t[p].l + t[p].r) >> 1, now = 0;
if (l0 <= mid) now = now | query(ls(p), l0, r0);
if (r0 > mid) now = now | query(rs(p), l0, r0);
return now;
}
int main()
{
n = read(); mx = read(); m = read();
build(1, 1, n);
while (m -- )
{
scanf("%s", ch);
if (ch[0] == 'C')
{
int l0 = read(), r0 = read(), c = read();
if (l0 > r0) swap(l0, r0);
update(1, l0, r0, c);
}
else
{
int l0 = read(), r0 = read();
if (l0 > r0) swap(l0, r0);
int cnt = query(1, l0, r0), res = 0;
for (int i = 0; i <= mx - 1; i ++ )
if (cnt & (1 << i))
res ++ ;
printf("%d
", res);
}
}
return 0;
}