BNUOJ 1260 Brackets Sequence

Brackets Sequence

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID: 1141
64-bit integer IO format: %lld      Java class name: Main

Special Judge

 

Let us define a regular brackets sequence in the following way: 

1. Empty sequence is a regular sequence. 
2. If S is a regular sequence, then (S) and [S] are both regular sequences. 
3. If A and B are regular sequences, then AB is a regular sequence. 

For example, all of the following sequences of characters are regular brackets sequences: 

(), [], (()), ([]), ()[], ()[()] 

And all of the following character sequences are not: 

(, [, ), )(, ([)], ([(] 

Some sequence of characters '(', ')', '[', and ']' is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

 

Input

The input file contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

 

Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

 

Sample Input

([(]

Sample Output

()[()]

Source

Northeastern Europe 2001

 

解题：这dp啊，我这种学渣啊，每做一次，就有一次新的感觉！水好深啊！

 

注意是单样例的！改成多样例，立马WA了

 

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <cstring>

 4 #include <cmath>

 5 #include <algorithm>

 6 #include <climits>

 7 #include <vector>

 8 #include <queue>

 9 #include <cstdlib>

10 #include <string>

11 #include <set>

12 #include <stack>

13 #define LL long long

14 #define pii pair<int,int>

15 #define INF 0x3f3f3f3f

16 using namespace std;

17 const int maxn = 110;

18 int dp[maxn][maxn],c[maxn][maxn] = {-1};

19 char str[maxn];

20 void print(int i,int j) {

21     if(i > j) return;

22     if(i == j) {

23         if(str[i] == '(' || str[j] == ')')

24             printf("()");

25         else printf("[]");

26     } else {

27         if(c[i][j] >= 0) {

28             print(i,c[i][j]);

29             print(c[i][j]+1,j);

30         } else {

31             if(str[i] == '(') {

32                 printf("(");

33                 print(i+1,j-1);

34                 printf(")");

35             } else {

36                 printf("[");

37                 print(i+1,j-1);

38                 printf("]");

39             }

40         }

41     }

42 }

43 void go() {

44     int len = strlen(str),i,j,k,theMin,t;

45     for(i = 0; i < len; i++) dp[i][i] = 1;

46     for(k = 1; k < len; k++) {

47         for(i = 0; i+k < len; i++) {

48             j = i+k;

49             theMin = dp[i][i]+dp[i+1][j];

50             c[i][j] = i;

51             for(t = i+1; t < j; t++) {

52                 if(dp[i][t]+dp[t+1][j] < theMin) {

53                     theMin = dp[i][t]+dp[t+1][j];

54                     c[i][j] = t;

55                 }

56             }

57             dp[i][j] = theMin;

58             if(str[i] == '(' && str[j] == ')' || str[i] == '[' && str[j] == ']') {

59                 if(dp[i+1][j-1] < theMin) {

60                     dp[i][j] = dp[i+1][j-1];

61                     c[i][j] = -1;

62                 }

63             }

64         }

65     }

66     print(0,len-1);

67 }

68 int main() {

69     scanf("%s",str);

70     go();

71     puts("");

72     return 0;

73 }

View Code