思路：用双向链表，next[x] = y表示颜色为x的珠子的下一个珠子的颜色是y，pre[x] = y表示表示颜色为x的珠子的上一个珠子的颜色是y。时间复杂度是O(n + m)
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1e5 + 5;
int next[maxn], pre[maxn];
int n, m, head, len;
//双向链表
void Insert(int key, int pos1, int pos2) {
++len;
next[key] = pos2;
next[pos1] = key;
pre[key] = pos1;
pre[pos2] = key;
}
void init(int n) {
head = 0; next[head] = 0; pre[head] = 0;
int x, prev = 0;
for(int i = 0; i < n; ++i) {
scanf("%d", &x);
Insert(x, prev, next[prev]);
prev = x;
}
}
void Delete(int key, int pos1, int pos2) {
--len;
next[pos1] = pos2;
pre[pos2] = pos1;
}
void print() {
printf("%d\n", len);
int nex = next[head];
while(nex) {
printf("%d ", nex);
nex = next[nex];
}
printf("\n");
}
int main() {
while(scanf("%d%d", &n, &m) == 2) {
len = 0;
init(n);
char op[5];
int x, y;
for(int i = 0; i < m; ++i) {
scanf("%s", op);
if(op[0] == 'A') {
scanf("%d%d", &x, &y);
Insert(y, pre[x], x);
}
else {
scanf("%d", &x);
Delete(x, pre[x], next[x]);
}
}
print();
}
return 0;
}如有不当之处欢迎指出！
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