括号匹配，结果为匹配后的串
一、记忆化搜索(search with finger)：
从[0, strlen(s)-1] 开始搜索，缩短区间，期间记录搜索过的值，防止重复
cut[l][r]代表区间[l, r]中需要从哪里分割，为输出
串长不超过100 复杂度可以接受
1 // Brackets sequence
2 // Search with fingers
3
4 #include <iostream>
5 #include <string>
6 #include <algorithm>
7 #include <cstring>
8 #include <cstdio>
9
10 #define MAXN 100
11
12 int n, dp[MAXN+10][MAXN+10], cut[MAXN+10][MAXN+10];
13 char s[111];
14
15 int SearchWithFinger(int l, int r)
16 {
17 if(l > r)
18 return dp[l][r] = 0;
19 if(l == r)
20 return dp[l][r] = 1;
21 if(dp[l][r]!=-1)
22 return dp[l][r];
23
24 int num = 1<<30;
25 if( (s[l]==‘(‘&&s[r]==‘)‘)||(s[l]==‘[‘&&s[r]==‘]‘) )
26 num = std::min(num, SearchWithFinger(l+1, r-1));
27 else if(s[l]==‘(‘||s[l]==‘[‘)
28 num = std::min(num, SearchWithFinger(l+1, r)+1);
29 else if(s[r]==‘)‘||s[r]==‘]‘)
30 num = std::min(num, SearchWithFinger(l, r-1)+1);
31
32 for(int m = l; m < r; ++m){
33 int temp = SearchWithFinger(l, m) + SearchWithFinger(m+1, r);
34 if(num >= temp){
35 num = temp;
36 cut[l][r] = m;
37 }
38 }
39
40 return dp[l][r] = num;
41
42 }
43
44 void PrintOut(int l, int r)
45 {
46 if(l > r)
47 return;
48 if(l == r){
49 if(s[l]==‘(‘||s[l]==‘)‘)
50 printf("()");
51 if(s[l]==‘[‘||s[l]==‘]‘)
52 printf("[]");
53 }
54 else{
55 if(cut[l][r] >= 0){
56 PrintOut(l, cut[l][r]);
57 PrintOut(cut[l][r]+1, r);
58 }
59 else{
60 if(s[l]==‘(‘ ){
61 printf("(");
62 PrintOut(l+1, r-1);
63 printf(")");
64 }
65 if(s[l]==‘[‘){
66 printf("[");
67 PrintOut(l+1, r-1);
68 printf("]");
69 }
70 }
71 }
72
73 }
74
75 int main(int argc, char const *argv[])
76 {
77 // freopen("in", "r", stdin);
78 // freopen("out", "w", stdout);
79 scanf("%d", &n);
80 getchar();
81 while(n--){
82
83 getchar();
84 gets(s);
85
86
87 memset(dp, -1, sizeof(dp));
88 memset(cut, -1, sizeof(cut));
89
90 SearchWithFinger(0, strlen(s)-1);
91
92 PrintOut(0, strlen(s)-1);
93
94 printf("\n");
95 if(n)
96 printf("\n");
97 }
98 return 0;
99 }
二、区间DP
dp[l][r]代表从区间[l, r]的串需要添加至少多少个括号才能匹配
枚举区间长度len ： 1 to strlen(s)-1
dp[i][i] = 0 任何一个单独的括号， 都需要添加至少一个括号才能匹配
cut[l][r]意义同上
1 // Brackets sequence
2 // dynamic programming
3
4 #include <cstdio>
5 #include <cstring>
6 #include <string>
7 #include <algorithm>
8 #include <iostream>
9
10 #define MAXN 100
11
12 int n, dp[MAXN+5][MAXN+5], cut[MAXN+5][MAXN+5];
13 char s[MAXN+10];
14
15 void PrintOut(int l, int r)
16 {
17 if(l > r)
18 return ;
19 if(l == r){
20 if(s[l]==‘(‘||s[l]==‘)‘)
21 printf("()");
22 if(s[l]==‘[‘||s[l]==‘]‘)
23 printf("[]");
24 }
25 else{
26 if(cut[l][r] >= 0){
27 PrintOut(l, cut[l][r]);
28 PrintOut(cut[l][r]+1, r);
29 }
30 else{
31 if(s[l]==‘(‘ ){
32 printf("(");
33 PrintOut(l+1, r-1);
34 printf(")");
35 }
36 if(s[l]==‘[‘){
37 printf("[");
38 PrintOut(l+1, r-1);
39 printf("]");
40 }
41 }
42 }
43
44 }
45
46 int main(int argc, char const *argv[])
47 {
48 // freopen("in", "r", stdin);
49
50 while(gets(s)){
51
52 memset(dp, 0, sizeof(dp));
53 memset(cut, -1, sizeof(cut));
54
55 for(int i = 0; i < strlen(s); ++i)
56 dp[i][i] = 1;
57
58 for(int len = 1; len < strlen(s); ++len){
59 for(int l = 0; l+len < strlen(s); ++l){
60 int r = l+len;
61 dp[l][r] = 1<<30;
62 if( (s[l]==‘(‘&&s[r]==‘)‘)||(s[l]==‘[‘&&s[r]==‘]‘) )
63 dp[l][r] = std::min(dp[l][r], dp[l+1][r-1]);
64 else if(s[l]==‘(‘||s[l]==‘[‘)
65 dp[l][r] = std::min(dp[l][r], dp[l+1][r]+1);
66 else if(s[r]==‘)‘||s[r]==‘]‘)
67 dp[l][r] = std::min(dp[l][r], dp[l][r-1]+1);
68 for(int m = l; m < r; ++m){
69 if(dp[l][r] >= dp[l][m] + dp[m+1][r]){
70 dp[l][r] = dp[l][m] + dp[m+1][r];
71 cut[l][r] = m;
72 }
73 }
74 }
75 }
76 // printf("%d\n", dp[0][strlen(s)-1]);
77 PrintOut(0, strlen(s)-1);
78 printf("\n");
79 }
80 return 0;
81 }
时间： 2024-10-05 04:30:56