一种基于bitset的优化（P8796 [蓝桥杯 2022 国 AC] 替换字符题解）自留 

 
1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 inline int rd(){
 5     int xx = 0, flag = 1; char ch = getchar();
 6     while(ch < '0' || ch > '9'){
 7         if(ch == '-')flag = -1;
 8         ch = getchar();
 9     }
10     while(ch >= '0' && ch <= '9'){
11         xx = xx * 10 + ch - '0';
12         ch = getchar();
13     }
14     return xx * flag;
15 }
16 
17 const int N = 1e5 + 5;
18 int n, m;
19 char s[N];
20 int L[400], R[400], belong[N], cnt = 0, length;
21 bitset<400>b[30][400];
22 void change(int l, int r, int f, int t){
23     int pl = belong[l], pr = belong[r];
24     if(pl == pr){
25         for(int i = l,j = l - L[pl] + 1; i <= r; i++, j++){
26             if(b[f][pl].test(j)){
27                 b[t][pl].set(j);
28                 b[f][pl].reset(j);
29             }
30         }
31     }else {
32         for(int i = l, j = l - L[pl] + 1; i <= R[pl]; i++, j++){
33             if(b[f][pl].test(j)){
34                 b[t][pl].set(j);
35                 b[f][pl].reset(j);
36             }
37         }
38         for(int i = L[pr], j = 1; i <= r; i++, j++){
39             if(b[f][pr].test(j)){
40                 b[t][pr].set(j);
41                 b[f][pr].reset(j);
42             }
43         }
44         
45         for(int i = pl + 1; i < pr; i++){
46             b[t][i] |= b[f][i];
47             b[f][i].reset();    
48         }
49     }
50 }
51 void print(){
52     memset(s, 0, sizeof(s));
53     for(int j = 1; j <= cnt; j++){
54         for(int k = L[j], o = 1; k <= R[j]; k++, o++){
55             for(int i = 1; i <= 26; i++){    
56                 if(b[i][j].test(o)){
57                     s[k] = i + 'a' - 1;
58                     break;
59                 } 
60             }
61         }
62     }
63     puts(s + 1);
64 }
65 int main(){
66     scanf("%s", s + 1);
67     n = strlen(s + 1);
68     length = pow(n, 0.5);
69     while(R[cnt] < n){
70         cnt ++;
71         L[cnt] = R[cnt - 1] + 1;
72         R[cnt] = L[cnt] + length - 1;
73     } R[cnt] = n;
74     for(int i = 1; i <= cnt; i++){
75         for(int j = L[i], k = 1; j <= R[i]; j++, k++){
76             belong[j] = i;
77             b[s[j] - 'a' + 1][i].set(k);
78         }    
79     }
80     //print();
81     m = rd();
82     int l, r, c1, c2;
83     for(int i = 1; i <= m; i++){
84         l = rd(), r = rd();
85         c1 = getchar() - 'a' + 1, getchar(), c2 = getchar() - 'a' + 1;
86         change(l, r, c1, c2);//print();
87     }          
88     print();
89 }

使用bitset可以快速实现复制，修改等操作，并且时间复杂度低。

posted @ 2025-05-18 17:30  指挥的智慧  阅读(12)  评论(0)    收藏  举报