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

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 int dp[maxn][maxn],c[maxn][maxn] = {-};
 char str[maxn];
 void print(int i,int j) {
     if(i > j) return;
     if(i == j) {
         if(str[i] == '(' || str[j] == ')')
             printf("()");
         else printf("[]");
     } else {
         if(c[i][j] >= ) {
             print(i,c[i][j]);
             print(c[i][j]+,j);
         } else {
             if(str[i] == '(') {
                 printf("(");
                 print(i+,j-);
                 printf(")");
             } else {
                 printf("[");
                 print(i+,j-);
                 printf("]");
             }
         }
     }
 }
 void go() {
     int len = strlen(str),i,j,k,theMin,t;
     for(i = ; i < len; i++) dp[i][i] = ;
     for(k = ; k < len; k++) {
         for(i = ; i+k < len; i++) {
             j = i+k;
             theMin = dp[i][i]+dp[i+][j];
             c[i][j] = i;
             for(t = i+; t < j; t++) {
                 if(dp[i][t]+dp[t+][j] < theMin) {
                     theMin = dp[i][t]+dp[t+][j];
                     c[i][j] = t;
                 }
             }
             dp[i][j] = theMin;
             if(str[i] == '(' && str[j] == ')' || str[i] == '[' && str[j] == ']') {
                 if(dp[i+][j-] < theMin) {
                     dp[i][j] = dp[i+][j-];
                     c[i][j] = -;
                 }
             }
         }
     }
     print(,len-);
 }
 int main() {
     scanf("%s",str);
     go();
     puts("");
     return ;
 }