NOIP2023模拟16联测37 D. 小猫吃火龙果
题目大意
有 \(n\) 个物品 \(A\) , \(B\) , \(C\) ，\(A\) 吃 \(B\)，\(B\) 吃 \(C\)，\(C\) 吃 \(A\)，有两种操作，给 \([ l , r ]\) 的 \(x , y\) 互换，求出经过操作后得出什么。
\(n , m \le 2\times10^5\)
思路
分块
维护一个状态 \(c_i\) 表示这个块的 \(A , B , C\) 分别变成了什么。
再维护一个 \(to_{id , x , y}\) 第 \(id\) 块，状态为 \(x\)，把 \(y\) 带进去会带出来什么。
code
#include <bits/stdc++.h>
#define fu(x , y , z) for(int x = y ; x <= z ; x ++)
#define get_next(now , x) (now == (x + 1) % 3 ? x : now)
using namespace std;
const int N = 2e5 + 5;
int n , m , a[N] , block , to[N][6][3] , t[N];
int c[6][3] = {{0 , 1 , 2} , {0 , 2 , 1} , {1 , 0 , 2} , {1 , 2 , 0} , {2 , 0 , 1} , {2 , 1 , 0}};
int b[6][3] = {{2 , 5 , 1} , {3 , 4 , 0} , {0 , 3 , 4} , {1 , 2 , 5} , {5 , 1 , 2} , {4 , 0 , 3}};
inline int read () {
char ch = getchar ();
while (ch != 'A' && ch != 'B' && ch != 'C') ch = getchar ();
return ch - 'A';
}
// inline int get_next (int now , int x) {
// if (now == (x + 1) % 3) return x;
// else return now;
// }
inline void updata (int id) {
int l = id * block + 1 , r = min ((id + 1) * block , n);
fu (i , 0 , 5) {
fu (j , 0 , 2) {
to[id][i][j] = j;
fu (k , l , r)
to[id][i][j] = get_next (to[id][i][j] , c[i][a[k]]);
}
}
}
inline void renew (int id) {
int l = id * block + 1 , r = min ((id + 1) * block , n);
fu (i , l , r) a[i] = c[t[id]][a[i]];
t[id] = 0;
}
inline int rd () {
int val = 0;
char ch = getchar ();
while (ch < '0' || ch > '9') ch = getchar ();
while (ch >= '0' && ch <= '9') {
val = val * 10 + (ch - '0');
ch = getchar ();
}
return val;
}
int main () {
freopen ("training.in" , "r" , stdin);
freopen ("training.out" , "w" , stdout);
int x , y , l , r , op , minr , id , type;
n = rd () , m = rd ();
fu (i , 1 , n) a[i] = read ();
if(n>100)block = sqrt (n / 12);
else block=sqrt(n);
fu (i , 0 , (n + block - 1) / block)
updata (i);
while (m --) {
op = rd () , l = rd () , r = rd ();
if (op) {
x = read ();
id = (l - 1) / block;
minr = min ((l - 1) / block * block + block , r);
while (l <= minr) {
x = get_next (x , c[t[id]][a[l]]);
l ++;
}
while (l <= r - block + 1) {
id = (l - 1) / block;
x = to[id][t[id]][x];
l += block;
}
id = (l - 1) / block;
while (l <= r) {
x = get_next (x , c[t[id]][a[l]]);
l ++;
}
printf ("%c\n" , x + 'A');
}
else {
x = read () , y = read ();
if (x == y) continue;
if (x > y) swap (x , y);
id = (l - 1) / block;
minr = min ((l - 1) / block * block + block , r);
type = (x == 0 ? y == 1 ? 0 : 1 : 2);
renew (id);
while (l <= minr) {
if (a[l] == x) a[l] = y;
else if (a[l] == y) a[l] = x;
l ++;
}
updata (id);
while (l <= r- block + 1) {
id = (l - 1) / block;
t[id] = b[t[id]][type];
l += block;
}
renew (id = (l - 1) / block);
while (l <= r) {
if (a[l] == x) a[l] = y;
else if (a[l] == y) a[l] = x;
l ++;
}
updata (id);
}
}
return 0;
}