数字替换
思路
由于每次替换不是永久性替换,因此我们可以考虑逆序去执行操作,这样每次的替换操作就可以当做是一个永久性替换
考虑用 \(f_i\) 表示 \(i\) 被替换后的数字,倒序遍历所有操作然后插入到答案数组即可
CODE
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define ll long long
#define rep(i, a, b) for(int i(a); i <= b; ++ i)
#define dec(i, a, b) for(int i(a); i >= b; -- i)
using namespace std;
const int N = 5e5 + 10;
namespace fast_IO {
char buf[N], *s, *t;
char getchar() {
return (s == t) && (t = (s = buf) + fread(buf, 1, N, stdin)), s == t ? -1 : *s++;
}
int read() {
int num = 0;
char c;
while ((c = getchar()) != '-' && c != '+' && (c < '0' || c > '9') && ~c);
num = c ^ 48;
while ((c = getchar()) >= '0' && c <= '9')
num = (num << 1) + (num << 3) + (c ^ 48);
return num;
}
}
struct node {
int op, x, y;
}q[N];
int f[N];
int main() {
int n = fast_IO::read();
int m = 0;
rep(i, 1, n) {
int op = fast_IO::read(); int x, y;
if(op & 1) {
++ m;
x = fast_IO::read();
} else {
x = fast_IO::read(), y = fast_IO::read();
}
q[i] = {op, x, y};
}
reverse(q + 1, q + n + 1);
rep(i, 1, N - 1) f[i] = i;
vector<int> ans(m); int id = 0;
rep(i, 1, n) {
int op = q[i].op, x = q[i].x, y = q[i].y;
if(op & 1) ans[id ++] = f[x];
else f[x] = f[y];
}
reverse(ans.begin(), ans.end());
for(int x: ans) printf("%d ", x);
return 0;
}